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Chapter 9: Problem 81

A gaseous mixture containing \(10.079 \mathrm{~g}\) of \(\mathrm{H}_{2}\) and \(7.00\) moles of \(\mathrm{Br}_{2}\) react to form HBr. (a) Write a balanced chemical equation for this reaction. (b) Which reactant is limiting? (c) What is the theoretical yield for this reaction in moles? (d) What is the theoretical yield for this reaction in grams? (e) How many moles of excess reactant are left over at the end of the reaction? (f) How many grams of excess reactant are left over at the end of the reaction?

Short Answer

Expert verified
(a) The balanced chemical equation is: \[ \mathrm{H}_{2} + \mathrm{Br}_{2} \to 2 \mathrm{HBr} \] (b) H2 is the limiting reactant. (c) The theoretical yield for this reaction in moles is 9.98 moles of HBr. (d) The theoretical yield for this reaction in grams is 807.6 g of HBr. (e) There are 2.01 moles of excess Br2 left over. (f) There are 321.2 g of excess Br2 left over.

Step by step solution

01

Write a balanced chemical equation for this reaction.

The balanced chemical equation for the reaction is: \[ \mathrm{H}_{2} + \mathrm{Br}_{2} \to 2 \mathrm{HBr} \]
02

Convert the given mass of hydrogen to moles.

To find the moles of hydrogen gas (H2), we use its molar mass, which is 2.02 g/mol. \[ \text{Moles of H}_{2} = \frac{10.079 \mathrm{~g}}{2.02 \mathrm{~g/mol}} = 4.99 \mathrm{~moles} \]
03

Determine the limiting reactant.

To determine the limiting reactant, we need to check which reactant runs out first in the reaction by comparing the moles of each reactant. According to the balanced equation, one mole of H2 reacts with one mole of Br2. \[ \text{Mole ratio} = \frac{\mathrm{Moles~of~H}_{2}}{\mathrm{Moles~of~Br}_{2}} = \frac{4.99}{7.00} = 0.71 \] Since the mole ratio is less than 1, H2 is the limiting reactant.
04

Calculate the theoretical yield in moles.

Using the mole ratio from the balanced equation, for every mole of H2, we get 2 moles of HBr. So: \[ \text{Theoretical yield in moles} = 4.99 \mathrm{~moles~of~H}_{2} \times 2 = 9.98 \mathrm{~moles~of~HBr} \]
05

Calculate the theoretical yield in grams.

To find the theoretical yield in grams, we convert moles of HBr into grams by using its molar mass, which is 80.9 g/mol. \[ \text{Theoretical yield in grams} = 9.98 \mathrm{~moles} \times 80.9 \mathrm{~g/mol} = 807.6 \mathrm{~g~of~HBr} \]
06

Calculate the moles of excess reactant left over.

Since we used 4.99 moles of H2, we calculate the moles of Br2 left over by subtracting the moles of Br2 consumed in the reaction. \[ \text{Moles of excess Br}_{2} = \text{Initial} - \text{Consumed} \] \[ \text{Moles of excess Br}_{2} = 7.00 - 4.99 = 2.01 \mathrm{~moles~of~Br}_{2} \]
07

Calculate the grams of excess reactant left over.

To find the leftover mass of Br2, we convert moles of Br2 into grams by using its molar mass, which is 159.8 g/mol. \[ \text{Grams of excess Br}_{2} = 2.01 \mathrm{~moles} \times 159.8 \mathrm{~g/mol} = 321.2 \mathrm{~g~of~Br}_{2} \] So, the answers for each part of this exercise are: (a) The balanced chemical equation is: \[ \mathrm{H}_{2} + \mathrm{Br}_{2} \to 2 \mathrm{HBr} \] (b) H2 is the limiting reactant. (c) The theoretical yield for this reaction in moles is 9.98 moles of HBr. (d) The theoretical yield for this reaction in grams is 807.6 g of HBr. (e) There are 2.01 moles of excess Br2 left over. (f) There are 321.2 g of excess Br2 left over.

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