Answer
7.45 g of $P_4$
Work Step by Step
1. Identify the limiting reactant.
- Calculate or find the molar mass for $ P_4 $:
$ P_4 $ : ( 30.97 $\times$ 4 )= 123.88 g/mol
- Using the molar mass as a conversion factor, find the amount in moles:
$$ 45.69 \space g \times \frac{1 \space mole}{ 123.88 \space g} = 0.3688 \space mole$$
- 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:
$$ 131.3 \space g \times \frac{1 \space mole}{ 70.90 \space g} = 1.852 \space moles$$
Find the amount of product if each reactant is completely consumed.
$$ 0.3688 \space mole \space P_4 \times \frac{ 4 \space moles \ PCl_3 }{ 1 \space mole \space P_4 } = 1.475 \space moles \space PCl_3 $$
$$ 1.852 \space moles \space Cl_2 \times \frac{ 4 \space moles \ PCl_3 }{ 6 \space moles \space Cl_2 } = 1.235 \space moles \space PCl_3 $$
Since the reaction of $ Cl_2 $ produces less $ PCl_3 $ for these quantities, it is the limiting reactant.
2. Find the amount of $ P_4 $ consumed:
$$ 1.852 \space moles \space Cl_2 \times \frac{ 1 \space mol \space P_4 }{ 6 \space mol \space Cl_2 } = 0.3087 \space mol \space P_4 $$
3. Subtract that amount from the amount in the reaction mixture.
$$Excess = 0.3688 - 0.3087 = 0.0601 \space mol \space P_4 $$
4. Using its molar mass, find the amount in grams:
$$ 0.0601 \space mole \times \frac{ 123.88 \space g}{1 \space mole} = 7.45 \space g$$
