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
$2.29 mol \times \frac{ 253.8 g}{ 1 mol} = 581g (I_2)$
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(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
$14.4 mol \times \frac{ 127.9 g}{ 1 mol} = 1847g (HI)$