Stoichiometry of Gases — Overview & Examples

Let's think about chemical reactions. Typically, we know the quantity of the reactants. But, we need to find the amount of product. We've already learned the tool for handling chemical reactions. It's stoichiometry! Now that we've learned about the gas laws, we can apply stoichiometry to gases.

Steps in a Stoichiometry Problem

Let's recall the steps in a stoichiometry problem.

Step 1: Write the chemical equation.

Step 2: Balance the chemical equation.

Step 3: Do any needed conversions. You should have SI units.

Step 4: Find the moles of the reactant(s).

• For help with this step, look at the stoichiometry roadmap or the mole roadmap.

• Gas stoichiometry problems may require you to use the ideal gas law.

Step 5: Determine the limiting reagent.

Step 6: Find the moles of the product(s). Use the mole ratios.

Step 7: If the question asks for something other than moles of product, solve the question.

• Sometimes, you will need to solve for pressure, volume, or temperature. I recommend using the ideal gas law.

• If the product is a solid, you may have to find its mass.

Example Problems from Easy to Hard

1. A canister containing the noble gas argon is at standard temperature and pressure. How many moles of argon do we have if the canister's volume is 14.75 liters?

• At STP, one mole of gas has a volume of 22.4 liters. So, all we need to do is a conversion.

• $\frac{1.0 mols}{22.4L} \times \frac{14.75L}{X moles} \rightarrow X=\frac{14.75L\times 1 mol}{22.4} \rightarrow X=0.66 mols$

1. Hydrogen peroxide undergoes a decomposition reaction. The reaction produces oxygen gas and liquid water. How many mols of oxygen did we create if 500.0 ml of a 5.0 molar hydrogen peroxide solution decomposed?

Step 1: Write the equation.

• $H_{2}O_{2(aq)} \rightarrow H_{2}O_{(l)} + O_{2(g)}$

Step 2: Balance the equation.

• $2H_{2}O_{2(aq)} \rightarrow 2H_{2}O_{(l)} + O_{2(g)}$

Step 3: Do any required conversions.

• Convert milliltres to liters. $500.0 ml \times \frac{1L}{1000 ml}=0.5L$

Step 4: Find the moles of the reactant.

• In this problem, you convert concentration to moles. $5.0 \frac{mols}{L} \times 0.5L= 2.5 mols_{(H_{2}O_{2})}$

Step 5: Find the limiting reagent.

• Hydrogen peroxide is the only reagent. So, it's limiting.

Step 6: Find the moles of the product.

• $2.5 mols_{(H_{2}O_{2})} \times \frac {1mol_{(O_{2})}}{2 mols_{(H_{2}O_{2})}} \rightarrow 1.25 mols_{(O_{2})}$

1. In pipe welding, a welder combusts acetylene gas ($C_{2}H_{2})$ with oxygen. The IUPAC name for acetylene is ethyne. In organic chemistry, the -yne ending denotes a triple bond. The most common acetylene tanks use a flow rate of $0.7\frac{m^{3}}{hr}$ at atmospheric pressure and $15.0^{o}C$. Let's assume oxygen is in excess. After 1 hour, how many moles of ethyne did the welder use? How many moles of carbon dioxide did he produce?

Step 1: Write the equation.

• $C_{2}H_{2} + O_{2} \rightarrow CO_{2} + H_{2}O$

Step 2: Balance the equation.

• $2C_{2}H_{2} + 5O_{2} \rightarrow 4CO_{2} +2 H_{2}O$

Step 3: Do any needed conversion.

• Volume: $0.7m^{3} \times \frac{1000L}{1 m^{3}} = 700 L$

• Temperature: $15.0^{o}C + 273.15= 288.15 K$

Step 4: Find the moles of the reactant.

• Use the ideal gas law to find the moles of ethyne.

• $PV=nRT \rightarrow 1.0 atm \times 700 L= n\times 0.08206\frac{atm\times L}{mol\times K}\times288.15K$

• $n=\frac{ 700 atm \times L}{0.08206\frac{atm \times L}{mol\times K} \times 288.15K} \rightarrow n=29.6 mols_{(C_{2}H_{2})}$.

Step 5: Find the limiting reagent.

• Oxygen is in excess, so ethyne is the limiting reagent.

Step 6: Determine the moles of the product.

• $29.6 mols_{(C_{2}H_{2})} \times \frac{4 mols_{(CO_{2})}}{2 mols_{(C_{2}H_{2})}} \rightarrow 59.2 mols_{(CO_{2})}$

1. Hydrogen chloride gas and ammonia undergo a synthesis reaction. It forms solid ammonium chloride. Our container has a volume of 5.0 L, a temperature of $5^{o}C$, and a pressure of 450 torr. The mole fraction of hydrogen chloride is 0.27. Find the mass of ammonium chloride produced by the reaction.

Step 1: Write the chemical equation.

• $HCl_{(g)} + NH_{3(g)} \rightarrow NH_{4}Cl_{(s)}$

Step 2: Balance the equation.

• The equation is already balanced.

Step 3: Do the required conversions.

• Pressure: $450.0 torr\times \frac{ 1.0 atm}{760.0 torr}= 0.59 atm.$

• Temperature: $5^{o}C + 273.15= 278.15 K$

• Mol fractions: $X_{NH_{3}}= 1-X_{HCl}= 1-0.27= .73$

Step 4: Find the moles of the products.

• In this problem, you need to start with Dalton's law. After finding the partial pressures, use the ideal gas law.

• $P_{HCl}=P_{total}\times X_{HCl}= 0.59 atm \times 0.27= 0.16 atm$

• $P_{HCl}\times V=n_{HCl}RT \rightarrow 0.16atm\times5.0L=n\times 0.8206\frac{atm\times L}{mols\times K}\times278.15K$

• $n_{HCl}=0.035 mols.$

• $P_{NH_{3}}=P_{total}\times X_{NH_{3}}= 0.59 atm \times 0.73= 0.43 atm$

• $P_{NH_{3}}\times V=n_{NH_{3}}RT \rightarrow 0.43atm\times5.0L=n\times 0.8206\frac{atm\times L}{mols\times K}\times278.15K$

• $n_{NH_{3}}=0.096 mols.$

Step 5: Determine the limiting reagent.

• Here the mole ratio is 1:1. So whichever reactant has the least number of moles is the limiting reagent.

• HCl is the limiting reagent.

Step 6: Find the moles of the product.

• Again, the ratio is 1:1. So, 0.035 mols of HCl produces 0.035 mols of $NH_{4}Cl$.

Step 7: Solve for whatever the question asks.

• The question asks for the mass of ammonium chloride.

• The molar mass of $NH_{4}Cl$ is $53.49\frac{g}{mol}$.

• $0.035 mols_{(NH_{4}Cl)}\times\frac{53.49g}{mol}= 1.87 g_{(NH_{4}Cl)}$

(Video) Gas Stoichiometry Problems

by The Organic Chemistry Tutor

In this video, The Organic Chemistry Tutor works through six example problems! Each one is a variation of gas stoichiometry. Do you need to practice stoichiometry with gases? This video is a great resource!