Question 3.4.1: Identify the limiting reactant in the reaction of hydrogen a......
Identify the limiting reactant in the reaction of hydrogen and oxygen to form water if 61.0 g of O_{2} and 8.40 g of H_{2} are combined. Determine the amount (in grams) of excess reactant that remains after the reaction is complete.
You are asked to identify the limiting reactant and mass (in grams) of the excess reactant remaining after the reaction is complete.
You are given the mass of the reactants.
Step 1. Write a balanced chemical equation.
2 H_{2} + O_{2} → 2 H_{2}O
Step 2. Determine the limiting reactant by comparing the relative amounts of reactants available.
Calculate the amount (in mol) of one of the reactants needed and compare that value to the amount available.
amount of O_{2} needed: 8.40 g H_{2}\left(\frac{1\text{ mol H}_{2}}{2.016\text{ g}} \right) \left(\frac{1\text{ mol O}_{2}}{2\text{ mol H}_{2}} \right) = 2.08 mol O_{2}
amount of O_{2} available: 61.0 g O_{2} × \frac{1\text{ mol O}_{2}}{32.00\text{ g}} = 1.91 mol O_{2}
More O_{2} is needed (2.08 mol) than is available (1.91 mol), so O_{2} is the limiting reactant.
Alternatively, the amounts of reactants available can be compared to the stoichiometric ratio in the balanced equation in order to determine the limiting reactant.
\frac{4.17\text{ mol H}_{2}}{1.91\text{ mol O}_{2}}=\underset{ \text{ratio of reactants} \\ \text{(available) }}{\frac{2\text{ mol H}_{2}}{1\text{ mol O}_{2}}}>\underset{\text{ratio of reactants}\\ \text{(balanced equation)}}{\frac{2\text{ mol H}_{2}}{1\text{ mol O}_{2}}}
Here, the mole ratio of H_{2} to O_{2} available is greater than the mole ratio from the balanced chemical equation. Thus, hydrogen is in excess and oxygen is the limiting reactant.
Step 3. Use the amount of limiting reactant (O_{2}) available to calculate the amount of excess reactant (H_{2}) needed for complete reaction.
1.91 mol O_{2}\left(\frac{2\text{ mol H}_{2}}{1\text{ mol O}_{2}} \right) \left(\frac{2.016\text{ g}}{1\text{ mol H}_{2}} \right) = 7.70 g H_{2}
Step 4. Calculate the mass of excess reactant that remains when all the limiting reactant is consumed.
8.40 g H_{2} available – 7.70 g H_{2} consumed = 0.70 g H_{2} remains
Is your answer reasonable? More O_{2} is needed than is available, so it is the limiting reactant. There is H_{2} remaining after the reaction is complete because it is the excess reactant in the reaction