Answer
(a) 365 g
(b) 716 g
Work Step by Step
(a)
1. Calculate the number of moles of $H_2S$:
1.008* 2 + 32.07* 1 ) = 34.086g/mol
$49.2g \times \frac{1 mol}{ 34.086g} = 1.44mol (H_2S)$
According to the balanced equation:
The ratio of $H_2S$ to $I_2$ is 1 to 1:
$1.44 mol (H_2S) \times \frac{ 1 mol(I_2)}{ 1 mol (H_2S)} = 1.44mol (I_2)$
2. Calculate the mass of $I_2$:
126.9* 2 = 253.8g/mol
$1.44 mol \times \frac{ 253.8 g}{ 1 mol} = 365 g (I_2)$
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(b)
1. Calculate the number of moles of $H_2S$:
1.008* 2 + 32.07* 1 = 34.086g/mol
$95.4g \times \frac{1 mol}{ 34.086g} = 2.8mol (H_2S)$
According to the balanced equation:
The ratio of $H_2S$ to $HI$ is 1 to 2:
$2.8 mol (H_2S) \times \frac{ 2 mol(HI)}{ 1 mol (H_2S)} = 5.6mol (HI)$
2. Calculate the mass of $HI$:
1.008* 1 + 126.9* 1 = 127.908g/mol
$5.6 mol \times \frac{ 127.908 g}{ 1 mol} = 716g (HI)$
