Question 4.15: Limiting Reagent Suppose 12 g of C is mixed with 64 g of O2 ...
Limiting Reagent
Suppose 12 g of C is mixed with 64 g of \text{O}_{2} and the following reaction takes place:
\text{C(s)} + \text{O}_{2}\text{(g)} \longrightarrow \text{CO}_{2}\text{(g)}(a) Which reactant is the limiting reagent, and which reactant is in excess?
(b) How many grams of \text{CO}_{2} will be formed?
Strategy
Determine how many moles of each reactant are present initially. Because C and \text{O}_{2} react in a 1:1 molar ratio, the reactant present in the smaller molar amount is the limiting reagent and determines how many moles and, therefore, how many grams of \text{CO}_{2} can be formed.
(a) We use the molar mass of each reactant to calculate the number of moles of each compound present before reaction.
12\ \cancel{\text{g C}} \times \frac{1\ \text{mol C}}{12\ \cancel{\text{g C}}} = 1.0\ \text{mol C}\\ 64\ \cancel{\text{g O}_{2}} \times \frac{1\ \text{mol O}_{2}}{32\ \cancel{\text{g O}_{2}}} = 2.0\ \text{mol O}_{2}According to the balanced equation, reaction of one mole of C requires one mole of \text{O}_{2}. But two moles of \text{O}_{2} are present at the start of the reaction. Therefore, C is the limiting reagent and \text{O}_{2} is in excess.
(b) To calculate the number of grams of \text{CO}_{2} formed, we use the conversion factor 1 mol \text{CO}_{2} = 44 g \text{CO}_{2}.
12\ \cancel{\text{g C}}\times \frac{1\ \cancel{\text{mol C}}}{12\ \cancel{\text{g C}}} \times \frac{1\ \cancel{\text{mol CO}_{2}}}{1\ \cancel{\text{mol C}}} \times \frac{44\ \text{g CO}_{2}}{1\ \cancel{\text{mol CO}_{2}}} = 44\ \text{g CO}_{2}We can summarize these numbers in the following table. Note that as required by the law of conservation of mass, the sum of the masses of the material present after reaction is the same as the amount present before any reaction took place, namely 76 g of material.
| \hspace{105 pt} \pmb{\text{C} \qquad+\quad \text{O}_{2}\quad \longrightarrow \quad\text{CO}_{2}} | |||
| 0 | 64 g | 12 g | Before reaction |
| 0 | 2.0 mol | 1.0 mol | Before reaction |
| 1.0 mol | 1.0 mol | 0 | After reaction |
| 44.0 g | 32.0 g | 0 | After reaction |
■ Quick Check 4.15
Assume that 6.0 g of C and 2.1 g of \text{H}_{2} are mixed and react to form methane according to the following balanced equation:
(a) Which is the limiting reagent, and which reactant is in excess?
(b) How many grams of \text{CH}_{4} are produced in the reaction?