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Chapter 3: Problem 49

A \(1.562 \mathrm{g}\) sample of the alcohol \(\mathrm{CH}_{3} \mathrm{CHOHCH}_{2} \mathrm{CH}_{3}\) is burned in an excess of oxygen. What masses of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) should be obtained?

Short Answer

Expert verified
The masses of CO2 and H2O that should be obtained are approximately 3.43 g and 1.87 g respectively.

Step by step solution

01

Balancing the reaction

The combustion reaction of the alcohol CH3CHOHCH3 in oxygen would give carbon dioxide (CO2) and water (H2O). Therefore, the unbalanced reaction would be: CH3CHOHCH3 + O2 → CO2 + H2O. The balanced chemical equation is: 2CH3CHOHCH3 + 9O2 → 6CO2 + 8H2O.
02

Convert mass of alcohol to moles

From the periodic table, the molar mass of CH3CHOHCH3 (using average atomic masses) is approximately 60.095 g/mol. Therefore, the number of moles of CH3CHOHCH3 burned is \(1.562 \mathrm{g} / 60.095 \mathrm{g/mol} = 0.026 \mathrm{mol}\).
03

Using stoichiometry to find moles of products

From the balanced equation, it can be seen that for every 2 moles of CH3CHOHCH3 combusted, 6 moles of CO2 and 8 moles of H2O are produced. So, the moles of CO2 produced is \(0.026 \mathrm{mol} CH3CHOHCH3 × (6 \mathrm{mol} CO2 / 2 \mathrm{mol} CH3CHOHCH3) = 0.078 \mathrm{mol} CO2\). Likewise, the moles of H2O produced is \(0.026 \mathrm{mol} CH3CHOHCH3 × (8 \mathrm{mol} H2O / 2 \mathrm{mol} CH3CHOHCH3) = 0.104 \mathrm{mol} H2O\).
04

Convert moles of products to mass

Using the molar masses of CO2 and H2O (approximately 44.01 g/mol and 18.015 g/mol respectively), the mass of CO2 and H2O produced are \(0.078 \mathrm{mol} CO2 × 44.01 \mathrm{g/mol} = 3.43 \mathrm{g}\) and \(0.104 \mathrm{mol} H2O × 18.015 \mathrm{g/mol} = 1.87 \mathrm{g}\) respectively.

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