Answer
(a) $S_8$ is the limiting reactant.
(b) 67.5 g of $S_2Cl_2$
(c) 35.6 g of $Cl_2$
Work Step by Step
- Calculate or find the molar mass for $ S_8 $:
$ S_8 $ : ( 32.07 $\times$ 8 )= 256.56 g/mol
- Using the molar mass as a conversion factor, find the amount in moles:
$$ 32.0 \space g \times \frac{1 \space mole}{ 256.56 \space g} = 0.125 \space mole$$
- Calculate or find the molar mass for $ Cl_2 $:
$ Cl_2 $ : ( 35.45 $\times$ 2 )= 70.90 g/mol
- Using the molar mass as a conversion factor, find the amount in moles:
$$ 71.0 \space g \times \frac{1 \space mole}{ 70.90 \space g} = 1.00 \space mole$$
Find the amount of product if each reactant is completely consumed.
$$ 0.125 \space mole \space S_8 \times \frac{ 4 \space moles \ S_2Cl_2 }{ 1 \space mole \space S_8 } = 0.500 \space mole \space S_2Cl_2 $$
$$ 1.00 \space mole \space Cl_2 \times \frac{ 4 \space moles \ S_2Cl_2 }{ 4 \space moles \space Cl_2 } = 1.00 \space mole \space S_2Cl_2 $$
Since the reaction of $ S_8 $ produces less $ S_2Cl_2 $ for these quantities, it is the limiting reactant.
- Calculate or find the molar mass for $ S_2Cl_2 $:
$ S_2Cl_2 $ : ( 35.45 $\times$ 2 )+ ( 32.07 $\times$ 2 )= 135.04 g/mol
- Using the molar mass as a conversion factor, find the mass in g:
$$ 0.500 \space mole \times \frac{ 135.04 \space g}{1 \space mole} = 67.5 \space g$$
- Find the amount of $Cl_2$ consumed.
$$ 0.125 \space mole \space S_8 \times \frac{ 4 \space moles \ Cl_2 }{ 1 \space mole \space S_8 } = 0.500 \space mole \space Cl_2 $$
$$ 0.500 \space mole \times \frac{ 70.90 \space g}{1 \space mole} = 35.4 \space g$$
Excess = Initial - Consumed = 71.0 g - 35.4 g = 35.6 g
