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Chapter 8: Problem 81

When a certain amount of octane, \(\mathrm{C}_{8} \mathrm{H}_{18}\), is burnt completely, \(7.04 \mathrm{~g} \mathrm{CO}_{2}\) is formed. What is the mass of \(\mathrm{H}_{2} \mathrm{O}\) formed, simultaneously? (a) \(1.62 \mathrm{~g}\) (c) \(6.48 \mathrm{~g}\) (c) \(3.24 \mathrm{~g}\) (d) \(2.28 \mathrm{~g}\)

Short Answer

Expert verified
Answer: Approximately 3.24 grams of H2O are formed simultaneously.

Step by step solution

01

Write the balanced chemical equation for the combustion of octane

The balanced chemical equation for the complete combustion of octane is: C8H18 (l) + 12.5 O2 (g) -> 8 CO2 (g) + 9 H2O (l) In this equation, 1 mole of octane reacts to produce 8 moles of CO2 and 9 moles of H2O.
02

Calculate the moles of CO2 formed

Given mass of CO2 is 7.04 g. The molar mass of CO2 is (12.01 g/mol for C) + 2 * (16.00 g/mol for O) = 44.01 g/mol. To find the moles of CO2 formed, divide the given mass by the molar mass of CO2. moles of CO2 = (7.04 g) / (44.01 g/mol) ≈ 0.16 moles
03

Use stoichiometry to find the moles of H2O formed

The balanced chemical equation states that 8 moles of CO2 are produced simultaneously with 9 moles of H2O. So, to find the moles of H2O formed, we can set up the following relation based on the stoichiometry of the combustion reaction: moles of H2O = (moles of CO2) * (9 moles of H2O / 8 moles of CO2) moles of H2O = (0.16 moles) * (9/8) ≈ 0.18 moles
04

Calculate the mass of H2O formed

Now that we have the moles of H2O formed, we can multiply this by the molar mass of H2O to find the mass. The molar mass of H2O is (2 * 1.01 g/mol for H) + (16.00 g/mol for O) = 18.02 g/mol. mass of H2O = (0.18 moles) * (18.02 g/mol) ≈ 3.24 g So, the mass of H2O formed simultaneously is approximately 3.24 g, which corresponds to option (c).

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