Answer
40.2 g of $H_2S$
Work Step by Step
- Calculate or find the molar mass for $ CH_4 $:
$ CH_4 $ : ( 1.008 $\times$ 4 )+ ( 12.01 $\times$ 1 )= 16.04 g/mol
- Using the molar mass as a conversion factor, find the amount in moles:
$$ 35.8 \space g \times \frac{1 \space mole}{ 16.04 \space g} = 2.23 \space moles$$
- Calculate or find the molar mass for $ S $:
$ S $ : 32.07 g/mol
- Using the molar mass as a conversion factor, find the amount in moles:
$$ 75.5 \space g \times \frac{1 \space mole}{ 32.07 \space g} = 2.35 \space moles$$
Find the amount of product if each reactant is completely consumed.
$$ 2.23 \space moles \space CH_4 \times \frac{ 2 \space moles \ H_2S }{ 1 \space mole \space CH_4 } = 4.46 \space moles \space H_2S $$
$$ 2.35 \space moles \space S \times \frac{ 2 \space moles \ H_2S }{ 4 \space moles \space S } = 1.18 \space moles \space H_2S $$
Since the reaction of $ S $ produces less $ H_2S $ for these quantities, it is the limiting reactant.
- Calculate or find the molar mass for $ H_2S $:
$ H_2S $ : ( 1.008 $\times$ 2 )+ ( 32.07 $\times$ 1 )= 34.09 g/mol
- Using the molar mass as a conversion factor, find the mass in g:
$$ 1.18 \space mole \times \frac{ 34.09 \space g}{1 \space mole} = 40.2 \space g$$
