## Chemistry (7th Edition)

Published by Pearson

# Chapter 3 - Mass Relationships in Chemical Reactions - Worked Example - Page 87: 11

#### Answer

Percent yield: $62.2\%$

#### Work Step by Step

1. Calculate the number of moles of $C_2H_4$: 12.01* 2 + 1.008* 4 = 28.05g/mol $4.6g \times \frac{1 mol}{ 28.05g} = 0.164mol (C_2H_4)$ The given equation is already balanced; therefore: The ratio of $C_2H_4$ to $C_2H_6O$ is 1 to 1: $0.164 mol (C_2H_4) \times \frac{ 1 mol(C_2H_6O)}{ 1 mol (C_2H_4)} = 0.164mol (C_2H_6O)$ 2. Calculate the mass of $C_2H_6O$: 12.01* 2 + 1.008* 6 + 16* 1 = 46.07g/mol $0.164 mol \times \frac{ 46.07 g}{ 1 mol} = 7.56g (C_2H_6O)$ 3. Now, calculate the percent yield: $\frac{4.7g}{7.56g} \times 100\% = 62.2\%$

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