Question 3.20: Calculating Quantities in a Limiting-Reactant Problem: Mass ......

Plan The amounts of two reactants are given, which means this is a limiting-reactant problem. (a) To determine the mass of product formed, we must find the limiting reactant by calculating which of the given masses of reactant forms less nitrogen gas. As always, we first write the balanced equation. We convert the grams of each reactant to moles using that reactant’s molar mass and then use the molar ratio from the balanced equation to find the number of moles of N_2 each reactant forms. Next, we convert the lower amount of N_2 to mass (see the road map (Fig 3.20)). (b) To determine the mass of the excess reactant, we use the molar ratio to calculate the mass of excess reactant that is required to react with the given amount of the limiting reactant. We subtract that mass from the given amount of excess reactant; this difference is the mass of unreacted excess reactant. (c) We use the values based on the limiting reactant for the reaction table (Table 1).

Solution (a) Writing the balanced equation:

Finding the amount (mol) of N_2 from the amount (mol) of each reactant:

       \text{For }N_2H_4\text{: Amount (mol) of }N_2H_4  =  1.00 \times 10^2  \cancel{g  N_2H_4}  \times  \frac{1  \text{mol}  N_2H_4}{32.05  \cancel{g  N_2H_4}}  =  3.12  \text{mol}  N_2H_4
                \text{Amount (mol) of }N_2  =  3.12  \cancel{\text{mol}  N_2H_4}  \times  \frac{3  \text{mol}  N_2}{2  \cancel{\text{mol }N_2H_4}}  =  4.68  \text{mol}  N_2
       \text{For }N_2O_4\text{: Amount (mol) of }N_2O_4  =  1.00 \times 10^2  \cancel{g  N_2O_4}  \times  \frac{1  \text{mol}  N_2O_4}{92.02  \cancel{g  N_2O_4}}  =  2.17  \text{mol}  N_2O_4
                \text{Amount (mol) of }N_2  =  2.17  \cancel{\text{mol}  N_2O_4}  \times  \frac{3  \text{mol}  N_2}{1  \cancel{\text{mol }N_2O_4}}  =  6.51  \text{mol}  N_2

Thus, N_2H_4 is the limiting reactant because it yields less N_2. Converting from amount (mol) of N_2 to mass (g):

(b) Finding the mass (g) of N_2O_4 that reacts with 1.00×10² g of N_2H_4:

         \text{Mass (g) of }N_2O_4  =  1.00\times 10^2  \cancel{g  N_2H_4}  \times  \frac{1  \cancel{\text{mol}  N_2H_4}}{32.05  \cancel{g  N_2H_4}}  \times  \frac{1  \cancel{\text{mol}  N_2O_4}}{2  \cancel{\text{mol}  N_2H_4}}   \frac{92.02  g  N_2O_4}{1  \cancel{\text{mol }N_2O_4}}
                     =  144  g  N_2O_4

(c) With N_2H_4 as the limiting reactant, the reaction table is (Table 1)

Check There are more grams of N_2O_4 than N_2H_4, but there are fewer moles of N_2O_4 because its ℳ is much higher. Rounding for N_2H_4:  100  g  N_2H_4  ×  1  \text{mol}/32  g  ≈  3  \text{mol};  ∼3  \text{mol}  ×  \frac{3}{2}  ≈  4.5  \text{mol}  N_2;  ∼4.5  \text{mol}  ×  30  \text{g/mol}  ≈  135  g  N_2.

Comment 1. Recall this common mistake in solving limiting-reactant problems: The limiting reactant is not the reactant present in fewer moles (or grams). Rather, it is the reactant that forms fewer moles (or grams) of product.
2. An alternative approach to finding the limiting reactant compares “How much is needed?” with “How much is given?” That is, based on the balanced equation,
• Find the amount (mol) of each reactant needed to react with the other reactant.
• Compare that needed amount with the given amount in the problem statement. There will be more than enough of one reactant (excess) and less than enough of the other (limiting).

For example, the balanced equation for this problem shows that 2 mol of N_2H_4 reacts with 1 mol of N_2O_4. The amount (mol) of N_2O_4 needed to react with the given 3.12 mol of N_2H_4 is

The amount of N_2H_4 needed to react with the given 2.17 mol of N_2O_4 is

We are given 2.17 mol of N_2O_4, which is more than the 1.56 mol of N_2O_4 needed, and we are given 3.12 mol of N_2H_4, which is less than the 4.34 mol of N_2H_4 needed. Therefore, N_2H_4 is limiting, and N_2O_4 is in excess.