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Chapter 5: Problem 94

Propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\) burns in oxygen to produce carbon dioxide gas and water vapor. (a) Write a balanced equation for this reaction. (b) Calculate the number of liters of carbon dioxide measured at STP that could be produced from \(7.45 \mathrm{~g}\) of propane.

Short Answer

Expert verified
The balanced equation for this reaction is \(\mathrm{C}_{3} \mathrm{H}_{8} + 5\mathrm{O}_{2} \rightarrow 3\mathrm{CO}_{2} + 4\mathrm{H}_{2}\mathrm{O}\). The volume of carbon dioxide that could be produced from 7.45 g of propane is 11.36 L

Step by step solution

01

Write the unbalanced chemical equation

First, it is important to write the reaction of propane burning in oxygen, which results in carbon dioxide and water vapor. The initial, unbalanced, equation is: \(\mathrm{C}_{3} \mathrm{H}_{8} + \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O}\)
02

Balance the chemical equation

Here, we will balance the equation by providing the required number of molecules for each reactant and product. The balanced equation is: \(\mathrm{C}_{3} \mathrm{H}_{8} + 5\mathrm{O}_{2} \rightarrow 3\mathrm{CO}_{2} + 4\mathrm{H}_{2}\mathrm{O}\)
03

Calculate moles of propane

Before we can find the number of liters of \(\mathrm{CO}_{2}\), we need to find the number of moles of propane. This can be done using the formula: moles = mass / molar mass. Here, the molar mass of propane (\(\mathrm{C}_{3} \mathrm{H}_{8}\)) is 44.10 g/mol. Thus the number of moles of propane is \(7.45 g / 44.10 g/mol = 0.169 mol\)
04

Calculate moles of carbon dioxide

In the balanced chemical equation, it is clear that 1 mole of propane gives 3 moles of carbon dioxide. Therefore, the number of moles of \(\mathrm{CO}_{2}\) produced by \(0.169 mol\) of propane is \(0.169 mol * 3 = 0.507 mol\)
05

Calculate volume of carbon dioxide

Lastly, using the molar volume of a gas at STP which is 22.4L/mol, the volume of carbon dioxide can be calculated as \(0.507 mol * 22.4 L/mol = 11.36 L\)

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