## Chemistry (7th Edition)

Published by Pearson

# Chapter 3 - Mass Relationships in Chemical Reactions - Section Problems - Page 107: 68

#### Answer

(a) 581 g (b) 1847 g

#### Work Step by Step

(a) 1. Calculate the number of moles of $N_2H_4$: 14.01* 2 + 1.008* 4 = 32.052g/mol $36.7g \times \frac{1 mol}{ 32.052g} = 1.14\ mol (N_2H_4)$ According to the balanced equation: The ratio of $N_2H_4$ to $I_2$ is 1 to 2: $1.14 mol (N_2H_4) \times \frac{ 2 mol(I_2)}{ 1 mol (N_2H_4)} = 2.29mol (I_2)$ 2. Calculate the mass of $I_2$: 126.9* 2 = 253.8g/mol $2.29 mol \times \frac{ 253.8 g}{ 1 mol} = 581g (I_2)$ -------- (b) 1. Calculate the number of moles of $N_2H_4$: 14.01* 2 + 1.008* 4 = 32.052g/mol $116g \times \frac{1 mol}{ 32.052g} = 3.61mol (N_2H_4)$ According to the balanced equation: The ratio of $N_2H_4$ to $HI$ is 1 to 4: $3.61 mol (N_2H_4) \times \frac{ 4 mol(HI)}{ 1 mol (N_2H_4)} = 14.4mol (HI)$ 2. Calculate the mass of $HI$: 1.008* 1 + 126.9* 1 = 127.9g/mol $14.4 mol \times \frac{ 127.9 g}{ 1 mol} = 1847g (HI)$

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