Stoichiometry and Mole Ratios
Stoichiometry and Balanced Equations
Stoichiometry is a branch of chemistry that deals with the quantitative relationship between reactants and products in a Balanced Equation. Using the balanced chemical equation along with knowledge of atomic mass, molecular mass, the mole, Avogadro's number, and molar volume, we can calculate the amounts of substances involved in a reaction.
Law of Conservation of Mass: Stoichiometric calculations obey this law — total mass of reactants equals total mass of products
Law of Definite Proportions: Also obeyed — a given compound always contains its constituent elements in a fixed ratio by mass
Assumptions: All reactants are completely converted into products, and no side reactions occur
A balanced chemical equation enables several types of quantitative calculations between reactants and products. Each type connects different ways of expressing amount — mass, moles, volume, and number of particles.
Mass-Mass Relationship: Given the mass of one substance, calculate the mass of any other substance in the reaction
Mole-Mole Relationship: Given the moles of one substance, calculate the moles of another using the coefficients from the balanced equation
Mass-Volume Relationship: Given the mass of one substance, calculate the volume of another (or vice versa), using density to convert between mass and volume
Types of Stoichiometric Relationships
•
Mass ↔ Mass: convert given mass to moles, apply mole ratio, convert result back to mass
•
Mass ↔ Moles: convert given mass to moles (or vice versa), apply mole ratio for the other substance
•
Mass ↔ Volume: convert given mass to moles, apply mole ratio, convert to mass, then use density to find volume
Constructing Mole Ratios from Balanced Equations
The coefficients in a Balanced Equation represent the relative number of moles of each substance involved in the reaction. A Mole Ratio is the ratio of these coefficients and is used as a conversion factor to relate the amounts of any two substances in the reaction.
$$2KOH + H_2SO_4 \longrightarrow K_2SO_4 + 2H_2O$$
The coefficients (2, 1, 1, 2) tell us that 2 moles of KOH react with 1 mole of H2SO4 to produce 1 mole of K2SO4 and 2 moles of H2O
$2$=moles of KOH consumed(mol)
$1$=moles of H2SO4 consumed(mol)
$1$=moles of K2SO4 produced(mol)
$2$=moles of H2O produced(mol)
any pair of substances
→a mole ratio can be written as the ratio of their coefficients, e.g. KOH:K2SO4 = 2:1, KOH:H2O = 2:2 = 1:1
Reading Mole Ratios Directly: From the balanced equation, the coefficient ratio of KOH to K2SO4 is 2:1 — every 2 moles of KOH produce exactly 1 mole of K2SO4
Simplifying Ratios: Mole ratios can be simplified like fractions — the ratio KOH:H2O = 2:2 simplifies to 1:1, meaning equal moles
Cross-Substance Ratios: Ratios are not limited to reactant-product pairs — you can write a ratio between any two substances, such as K2SO4:H2O = 1:2
To construct and apply a Mole Ratio, follow a systematic approach: identify the given substance and the target substance from the balanced equation, write their coefficient ratio, then use it to convert moles of the given substance to moles of the target.
Step 1 — Identify Substances: Locate both the given substance and the target substance in the balanced equation
Step 2 — Write Coefficient Ratio: Extract the coefficients of both substances and write them as a ratio (target : given)
Step 3 — Multiply: Multiply the known moles of the given substance by this ratio to get the moles of the target substance
Mole Ratios from $2KOH + H_2SO_4 \rightarrow K_2SO_4 + 2H_2O$
•
KOH : K2SO4 = 2 : 1
•
KOH : H2O = 2 : 2 = 1 : 1
•
H2SO4 : K2SO4 = 1 : 1
•
H2SO4 : H2O = 1 : 2
•
K2SO4 : H2O = 1 : 2
Stoichiometric Calculations Using Balanced Equations
Most stoichiometric calculations follow a three-step process: convert the given quantity to moles, use the Mole Ratio from the balanced equation to find moles of the target substance, then convert the result to the desired unit. This framework works for mass-mass, mole-mole, and mass-volume calculations alike.
$$n = \frac{m}{M}$$
Converts between mass and moles — dividing mass by molar mass gives the number of moles
$n$=number of moles(mol)
$m$=mass of the substance(g)
$M$=molar mass of the substance(g/mol)
converting moles to number of particles
→multiply moles by Avogadro's number ($6.02 \times 10^{23}$ mol$^{-1}$)
converting moles back to mass
→multiply moles by molar mass: $m = n \times M$
Step 1 — Convert to Moles: If the given quantity is a mass, divide by molar mass to get moles. If already in moles, proceed directly to Step 2
Step 2 — Apply Mole Ratio: Use the coefficients from the balanced equation to convert moles of the given substance to moles of the target substance
Step 3 — Convert to Desired Unit: Convert the resulting moles to mass (× molar mass), volume (÷ density), or number of particles (× Avogadro's number)
Three-Step Stoichiometric Method
1
Given quantity → Moles (using molar mass or Avogadro's number)
2
Moles of given → Moles of target (using mole ratio from balanced equation)
3
Moles of target → Desired quantity (mass, volume, or particle count)
In a mass-mass calculation, the mass of one substance is given and the mass of another substance must be determined. This requires converting the given mass to moles, applying the Mole Ratio, and converting the resulting moles back to mass using the Molar Mass of the target substance.
$$m_{\text{target}} = m_{\text{given}} \times \frac{1}{M_{\text{given}}} \times \frac{c_{\text{target}}}{c_{\text{given}}} \times M_{\text{target}}$$
Combines all three steps into one expression: convert given mass to moles, apply mole ratio, convert back to mass
$m_{\text{given}}$=mass of the given substance(g)
$M_{\text{given}}$=molar mass of the given substance(g/mol)
$c_{\text{target}}$=coefficient of the target substance in the balanced equation(dimensionless)
$c_{\text{given}}$=coefficient of the given substance in the balanced equation(dimensionless)
$M_{\text{target}}$=molar mass of the target substance(g/mol)
$c_{\text{target}} = c_{\text{given}}$
→the mole ratio is 1:1 and cancels out — moles are equal, so only molar masses matter
Given Mass to Given Moles: Divide the given mass by the molar mass of the given substance
Given Moles to Target Moles: Multiply by the ratio (target coefficient / given coefficient)
Target Moles to Target Mass: Multiply the target moles by the molar mass of the target substance
When the target substance is a solution with known percentage concentration and density, additional conversion steps are needed after finding the moles of solute. The percentage by weight relates the mass of solute to the total mass of solution, and density relates the solution mass to its volume.
$$V = \frac{m_{\text{solute}}}{\% \times \rho}$$
Converts mass of solute to volume of solution by accounting for concentration (percentage) and density
$V$=volume of the solution(cm³)
$m_{\text{solute}}$=mass of the solute (pure substance)(g)
$\%$=percentage concentration by weight (expressed as a decimal, e.g. 27% = 0.27)(dimensionless)
$\rho$=density of the solution(g/cm³)
$\% = 1$ (100% pure substance)
→the formula reduces to $V = m / \rho$, the standard density formula
From Moles to Solute Mass: After finding moles of solute via mole ratio, multiply by its molar mass to get the mass of pure solute needed
From Solute Mass to Solution Mass: Divide the solute mass by the percentage concentration (as a decimal) to get the total mass of solution
From Solution Mass to Volume: Divide the solution mass by the density of the solution to obtain the volume required
Calculation Sequence for Solution-Based Problems
1
Given mass of product → moles of product (÷ molar mass of product)
2
Moles of product → moles of solute (× mole ratio from balanced equation)
3
Moles of solute → mass of solute (× molar mass of solute)
4
Mass of solute → mass of solution (÷ percentage concentration as decimal)
5
Mass of solution → volume of solution (÷ density of solution)
Application Strategy
Worked Examples
Problem Statement
Calculate the mass of K2SO4 and water produced when 14 g of KOH react with excess H2SO4. Also calculate the number of molecules of water produced.
mole ratiomass-mass calculationmole-mole calculation
Strategic Toolkit
$$2KOH + H_2SO_4 \longrightarrow K_2SO_4 + 2H_2O$$
Step 1
Convert given mass of KOH to moles
Find the molar mass of KOH (K = 39, O = 16, H = 1) and divide the given mass by it.
$$n_{KOH} = \frac{14.0 \text{ g}}{56 \text{ g/mol}} = 0.25 \text{ mol}$$
Step 2
Find moles of K2SO4 using mole ratio
From the balanced equation, the ratio KOH:K2SO4 = 2:1. Multiply the moles of KOH by (1/2) to get moles of K2SO4.
$$n_{K_2SO_4} = 0.25 \text{ mol} \times \frac{1}{2} = 0.125 \text{ mol}$$
Step 3
Convert moles of K2SO4 to mass
Multiply the moles of K2SO4 by its molar mass (2 × 39 + 32 + 4 × 16 = 174 g/mol).
$$m_{K_2SO_4} = 0.125 \text{ mol} \times 174 \text{ g/mol} = 21.75 \text{ g}$$
Step 4
Find moles of H2O using mole ratio
From the balanced equation, the ratio KOH:H2O = 2:2 = 1:1, so the moles of water equal the moles of KOH.
$$n_{H_2O} = 0.25 \text{ mol} \times \frac{2}{2} = 0.25 \text{ mol}$$
Step 5
Convert moles of H2O to mass
Multiply the moles of water by its molar mass (2 × 1 + 16 = 18 g/mol).
$$m_{H_2O} = 0.25 \text{ mol} \times 18 \text{ g/mol} = 4.50 \text{ g}$$
Step 6
Convert moles of H2O to number of molecules
Multiply the moles of water by Avogadro's number to find the total number of water molecules produced.
$$N = 0.25 \text{ mol} \times 6.02 \times 10^{23} \text{ mol}^{-1} = 1.505 \times 10^{23} \text{ molecules}$$
Problem Statement
Mg metal reacts with HCl to produce hydrogen gas. What is the minimum volume of HCl solution (27% by weight) required to produce 12.1 g of H2? The density of HCl solution is 1.14 g/cm³.
mole ratiomass-mass calculationmass-volume calculationpercentage concentration
Strategic Toolkit
$$Mg + 2HCl \longrightarrow MgCl_2 + H_2$$
Step 1
Convert given mass of H2 to moles
Use the molar mass of H2 (2 × 1.008 = 2.016 g/mol) to convert 12.1 g to moles.
$$n_{H_2} = \frac{12.1 \text{ g}}{2.016 \text{ g/mol}} = 6.0 \text{ mol}$$
Step 2
Find moles of HCl using mole ratio
From the balanced equation, the ratio H2:HCl = 1:2. Multiply the moles of H2 by (2/1) to get moles of HCl.
$$n_{HCl} = 6.0 \text{ mol} \times \frac{2}{1} = 12.0 \text{ mol}$$
Step 3
Convert moles of HCl to mass of HCl
Multiply the moles of HCl by its molar mass (1 + 35.5 = 36.5 g/mol) to find the mass of pure HCl required.
$$m_{HCl} = 12.0 \text{ mol} \times 36.5 \text{ g/mol} = 438 \text{ g}$$
Step 4
Convert mass of HCl to mass of HCl solution
The solution is 27% by weight, meaning 27 g HCl per 100 g solution. Divide the HCl mass by 0.27 to get the total solution mass.
$$m_{\text{solution}} = \frac{438 \text{ g}}{0.27} = 1622.2 \text{ g}$$
Step 5
Convert mass of solution to volume using density
Divide the total solution mass by the density (1.14 g/cm³) to find the volume of HCl solution needed.
$$V = \frac{1622.2 \text{ g}}{1.14 \text{ g/cm}^3} = 1423 \text{ cm}^3$$
Problem Statement
In the reaction $2KOH + H_2SO_4 \rightarrow K_2SO_4 + 2H_2O$, if 0.5 moles of H2SO4 are consumed, how many moles of K2SO4 and H2O are produced?
mole ratiomole-mole calculation
Strategic Toolkit
$$2KOH + H_2SO_4 \longrightarrow K_2SO_4 + 2H_2O$$
Step 1
Find moles of K2SO4 using mole ratio
From the balanced equation, the ratio H2SO4:K2SO4 = 1:1. Since the ratio is 1:1, the moles are equal.
$$n_{K_2SO_4} = 0.5 \text{ mol} \times \frac{1}{1} = 0.5 \text{ mol}$$
Step 2
Find moles of H2O using mole ratio
From the balanced equation, the ratio H2SO4:H2O = 1:2. Multiply 0.5 mol H2SO4 by 2 to get moles of water.
$$n_{H_2O} = 0.5 \text{ mol} \times \frac{2}{1} = 1.0 \text{ mol}$$
Problem Statement
How many grams of CO2 are produced when 16 g of CH4 undergoes complete combustion? ($CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$)
mole ratiomass-mass calculation
Strategic Toolkit
$$CH_4 + 2O_2 \longrightarrow CO_2 + 2H_2O$$
Step 1
Convert mass of CH4 to moles
Find the molar mass of CH4 (12 + 4 × 1 = 16 g/mol) and divide the given mass by it.
$$n_{CH_4} = \frac{16 \text{ g}}{16 \text{ g/mol}} = 1.0 \text{ mol}$$
Step 2
Find moles of CO2 using mole ratio
From the balanced equation, the ratio CH4:CO2 = 1:1. The moles of CO2 equal the moles of CH4.
$$n_{CO_2} = 1.0 \text{ mol} \times \frac{1}{1} = 1.0 \text{ mol}$$
Step 3
Convert moles of CO2 to mass
Multiply the moles of CO2 by its molar mass (12 + 2 × 16 = 44 g/mol).
$$m_{CO_2} = 1.0 \text{ mol} \times 44 \text{ g/mol} = 44 \text{ g}$$
Memorize formulas
Moles from Mass
Convert mass (g) to moles by dividing by molar mass (g/mol). This is the first step in all stoichiometric calculations where mass is given.
Formula
$$n = \frac{m}{M}$$
Mass from Moles
Convert moles to mass (g) by multiplying by molar mass (g/mol). Used to express the final answer in grams.
Formula
$$m = n \times M$$
Particles from Moles
Convert moles to number of molecules or atoms by multiplying by Avogadro's number ($6.02 \times 10^{23}$ per mole).
Formula
$$N = n \times N_A$$
Volume from Mass and Density
Convert mass of a solution to volume (cm³) by dividing by density (g/cm³). Used when working with liquid solutions.
Formula
$$V = \frac{m}{\rho}$$
Solution Volume from Solute Mass
Convert mass of pure solute to volume of solution by dividing by (percentage as decimal × density). Used for concentration-based problems.
Formula
$$V = \frac{m_{\text{solute}}}{\% \times \rho}$$
Review
1
Which of the following best describes the term stoichiometry?
Ans
The quantitative relationship between reactants and products in a balanced chemical equation
Review
Tap to reveal
Identify the definition
Stoichiometry combines the Greek words 'stoicheion' (element) and 'metron' (measure), referring to the measurement of elements in chemical reactions.
Stoichiometry = quantitative analysis of reactants and products
2
The coefficients in a balanced chemical equation directly represent:
Ans
The mole ratio of all reactants and products involved
Review
Tap to reveal
Read coefficients from balanced equation
For example, in $$2H_2 + O_2 \rightarrow 2H_2O$$, the coefficients 2, 1, and 2 give the mole ratio.
Coefficients = mole ratio
3
Given the balanced equation $$2H_2 + O_2 \rightarrow 2H_2O$$, what is the mole ratio of $H_2$ consumed to $H_2O$ produced?
Ans
1:1
Review
Tap to reveal
Identify the relevant coefficients
The coefficient of $H_2$ is 2 and the coefficient of $H_2O$ is 2.
$$H_2 : H_2O = 2 : 2$$
Simplify the ratio
Divide both sides by 2 to obtain the simplest whole-number ratio.
$$H_2 : H_2O = 1 : 1$$
4
Using the equation $$N_2 + 3H_2 \rightarrow 2NH_3$$, how many moles of $NH_3$ are produced when 6 moles of $H_2$ react with excess $N_2$?
Ans
4 moles
Review
Tap to reveal
Write the mole ratio from the equation
The ratio of $H_2$ to $NH_3$ is 3:2.
$$\frac{n(NH_3)}{n(H_2)} = \frac{2}{3}$$
Calculate moles of $NH_3$
Multiply the given moles of $H_2$ by the mole ratio.
$$n(NH_3) = 6 \times \frac{2}{3} = 4 \text{ mol}$$
5
Given the reaction $$2Na + Cl_2 \rightarrow 2NaCl$$, what mass of $NaCl$ is produced from 4.6 g of $Na$? (M: Na = 23 g/mol, NaCl = 58.5 g/mol)
Ans
11.7 g
Review
Tap to reveal
Convert mass of Na to moles
Use $$n = \frac{m}{M}$$ to find moles of sodium.
$$n(Na) = \frac{4.6}{23} = 0.2 \text{ mol}$$
Apply the mole ratio
From the balanced equation, 2 mol Na produces 2 mol NaCl, a 1:1 ratio.
$$n(NaCl) = 0.2 \text{ mol}$$
Convert moles of NaCl to mass
Use $$m = n \times M$$ to find the mass of NaCl produced.
$$m(NaCl) = 0.2 \times 58.5 = 11.7 \text{ g}$$
6
How many molecules are present in 0.5 moles of $CO_2$? ($N_A = 6.02 \times 10^{23}$ mol$^{-1}$)
Ans
$3.01 \times 10^{23}$
Review
Tap to reveal
Apply Avogadro's number formula
Use $$N = n \times N_A$$ with the given number of moles.
$$N = 0.5 \times 6.02 \times 10^{23}$$
Calculate the result
Multiply to find the total number of molecules.
$$N = 3.01 \times 10^{23} \text{ molecules}$$
7
Given the combustion reaction $$CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$$, what mass of $O_2$ is required to completely combust 8 g of $CH_4$? (M: $CH_4$ = 16 g/mol, $O_2$ = 32 g/mol)
Ans
32 g
Review
Tap to reveal
Convert mass of $CH_4$ to moles
Use $$n = \frac{m}{M}$$.
$$n(CH_4) = \frac{8}{16} = 0.5 \text{ mol}$$
Use mole ratio to find moles of $O_2$
From the balanced equation, 1 mol $CH_4$ requires 2 mol $O_2$.
$$n(O_2) = 0.5 \times 2 = 1.0 \text{ mol}$$
Convert moles of $O_2$ to mass
Use $$m = n \times M$$.
$$m(O_2) = 1.0 \times 32 = 32 \text{ g}$$
8
At STP, what volume is occupied by 2 moles of any gas?
Ans
44.8 dm³
Review
Tap to reveal
State the molar volume at STP
One mole of any gas occupies 22.4 dm³ at standard temperature and pressure.
$$V_m = 22.4 \text{ dm}^3\text{/mol}$$
Calculate total volume
Multiply the number of moles by the molar volume.
$$V = 2 \times 22.4 = 44.8 \text{ dm}^3$$
9
Given the balanced equation $$Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2$$, what is the mole ratio of $CO$ consumed to $Fe$ produced?
Ans
3:2
Review
Tap to reveal
Identify coefficients from the balanced equation
The coefficient of $CO$ is 3 and the coefficient of $Fe$ is 2.
$$CO : Fe = 3 : 2$$
10
Calculate the mass of 0.25 moles of $H_2SO_4$. (M: $H_2SO_4$ = 98 g/mol)
Ans
24.5 g
Review
Tap to reveal
Apply the mass-from-moles formula
Use $$m = n \times M$$ with the given moles and molar mass.
$$m = 0.25 \times 98$$
Calculate the result
Multiply to find the mass in grams.
$$m = 24.5 \text{ g}$$
11
Given the neutralization reaction $$NaOH + HCl \rightarrow NaCl + H_2O$$, if 20 g of $NaOH$ reacts completely with excess $HCl$, how many grams of $NaCl$ are produced? (M: NaOH = 40 g/mol, NaCl = 58.5 g/mol)
Ans
29.25 g
Review
Tap to reveal
Convert mass of NaOH to moles
Use $$n = \frac{m}{M}$$.
$$n(NaOH) = \frac{20}{40} = 0.5 \text{ mol}$$
Apply the mole ratio
From the balanced equation, 1 mol NaOH produces 1 mol NaCl (1:1 ratio).
$$n(NaCl) = 0.5 \text{ mol}$$
Convert moles of NaCl to mass
Use $$m = n \times M$$.
$$m(NaCl) = 0.5 \times 58.5 = 29.25 \text{ g}$$
12
The density of ethanol is 0.789 g/cm³. What volume is occupied by 15.78 g of ethanol?
Ans
20 cm³
Review
Tap to reveal
Apply the volume-from-mass formula
Use $$V = \frac{m}{\rho}$$ to calculate volume.
$$V = \frac{15.78}{0.789}$$
Calculate the result
Divide mass by density to obtain volume.
$$V = 20 \text{ cm}^3$$
13
In the balanced equation $$2H_2O_2 \rightarrow 2H_2O + O_2$$, how many moles of $O_2$ are produced when 4 moles of $H_2O_2$ decompose?
Ans
2 moles
Review
Tap to reveal
Write the mole ratio from the equation
The ratio of $H_2O_2$ to $O_2$ is 2:1.
$$\frac{n(O_2)}{n(H_2O_2)} = \frac{1}{2}$$
Calculate moles of $O_2$ produced
Multiply the given moles by the mole ratio.
$$n(O_2) = 4 \times \frac{1}{2} = 2 \text{ mol}$$
14
A 10% (w/w) $HCl$ solution has a density of 1.05 g/cm³. What volume of this solution contains 0.5 mol of $HCl$? (M: HCl = 36.5 g/mol)
Ans
173.8 cm³
Review
Tap to reveal
Convert moles of HCl to mass
Use $$m = n \times M$$ to find the mass of solute needed.
$$m(HCl) = 0.5 \times 36.5 = 18.25 \text{ g}$$
Apply the solution volume formula
Use $$V = \frac{m_{solute}}{\% \times \rho}$$ with w/w fraction and density.
$$V = \frac{18.25}{0.10 \times 1.05}$$
Calculate the volume
Divide to find the volume of solution required.
$$V = 173.8 \text{ cm}^3$$
15
Given the reaction $$2Mg + O_2 \rightarrow 2MgO$$, if 3 moles of $Mg$ react with 1 mole of $O_2$, which reactant is limiting and how many moles of $MgO$ are formed?
Ans
$O_2$ is limiting; 2 moles of MgO are formed
Review
Tap to reveal
Determine which reactant is limiting
The equation requires 2 mol Mg per 1 mol $O_2$. With 1 mol $O_2$, only 2 mol Mg can react, leaving 1 mol Mg unused.
$$O_2 \text{ is the limiting reactant}$$
Calculate moles of MgO from limiting reactant
From the equation, 1 mol $O_2$ produces 2 mol MgO.
$$n(MgO) = 1 \times \frac{2}{1} = 2 \text{ mol}$$
16
How many formula units are present in 0.2 moles of $Ca(NO_3)_2$? ($N_A = 6.02 \times 10^{23}$ mol$^{-1}$)
Ans
$1.204 \times 10^{23}$
Review
Tap to reveal
Apply Avogadro's number formula
Use $$N = n \times N_A$$ with the given number of moles.
$$N = 0.2 \times 6.02 \times 10^{23}$$
Calculate the result
Multiply to find the total number of formula units.
$$N = 1.204 \times 10^{23} \text{ formula units}$$
17
Given the reaction $$CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2$$, what volume of $CO_2$ at STP is produced from 25 g of $CaCO_3$? (M: $CaCO_3$ = 100 g/mol)
Ans
5.6 dm³
Review
Tap to reveal
Convert mass of $CaCO_3$ to moles
Use $$n = \frac{m}{M}$$.
$$n(CaCO_3) = \frac{25}{100} = 0.25 \text{ mol}$$
Apply mole ratio to find moles of $CO_2$
From the balanced equation, 1 mol $CaCO_3$ produces 1 mol $CO_2$.
$$n(CO_2) = 0.25 \text{ mol}$$
Convert moles of $CO_2$ to volume at STP
Use molar volume: $V = n \times 22.4$ dm³/mol.
$$V = 0.25 \times 22.4 = 5.6 \text{ dm}^3$$
18
In the reaction $$N_2 + 3H_2 \rightarrow 2NH_3$$ (all gases at the same temperature and pressure), what is the total volume of products compared to the total volume of reactants when 1 mole of $N_2$ reacts?
Ans
Products occupy half the volume of reactants
Review
Tap to reveal
Sum the coefficients of gaseous reactants
The reactant gas coefficients are 1 ($N_2$) and 3 ($H_2$).
$$V_{reactants} \propto 1 + 3 = 4 \text{ volumes}$$
Sum the coefficients of gaseous products
The product gas coefficient is 2 ($NH_3$).
$$V_{products} \propto 2 \text{ volumes}$$
Compare the volumes
The ratio of product volume to reactant volume.
$$\frac{V_{products}}{V_{reactants}} = \frac{2}{4} = \frac{1}{2}$$
19
Given the decomposition $$2KClO_3 \rightarrow 2KCl + 3O_2$$, if 67.2 dm³ of $O_2$ is collected at STP, what mass of $KClO_3$ decomposed? (M: $KClO_3$ = 122.5 g/mol)
Ans
245 g
Review
Tap to reveal
Convert $O_2$ volume to moles at STP
Use $n = V / 22.4$ dm³/mol.
$$n(O_2) = \frac{67.2}{22.4} = 3 \text{ mol}$$
Apply mole ratio to find moles of $KClO_3$
From the equation, 3 mol $O_2$ is produced by 2 mol $KClO_3$.
$$n(KClO_3) = 3 \times \frac{2}{3} = 2 \text{ mol}$$
Convert moles of $KClO_3$ to mass
Use $$m = n \times M$$.
$$m(KClO_3) = 2 \times 122.5 = 245 \text{ g}$$
20
What volume of a 20% (w/w) $NaOH$ solution (density = 1.22 g/cm³) contains 24.4 g of $NaOH$?
Ans
100 cm³
Review
Tap to reveal
Apply the solution volume formula
Use $$V = \frac{m_{solute}}{\% \times \rho}$$ with the given values.
$$V = \frac{24.4}{0.20 \times 1.22}$$
Calculate the result
Divide the mass of solute by the product of percentage and density.
$$V = \frac{24.4}{0.244} = 100 \text{ cm}^3$$
Question 1 of 20
0:00
1
Which of the following best describes the term stoichiometry?