Answer
a. 3.75 moles of $Cl_2$
b. 186 g of $AlCl_3$
c. 10.0 g of $AlCl_3$
d. 85.6 %
Work Step by Step
a. $$ 2.50 \space moles \space Al \times \frac{ 3 \space moles \ Cl_2 }{ 2 \space moles \space Al } = 3.75 \space moles \space Cl_2 $$
b. $ Al $ : 26.98 g/mol
$$ \frac{1 \space mole \space Al }{ 26.98 \space g \space Al } \space and \space \frac{ 26.98 \space g \space Al }{1 \space mole \space Al }$$
$ AlCl_3 $ : ( 26.98 $\times$ 1 )+ ( 35.45 $\times$ 3 )= 133.33 g/mol
$$ \frac{1 \space mole \space AlCl_3 }{ 133.33 \space g \space AlCl_3 } \space and \space \frac{ 133.33 \space g \space AlCl_3 }{1 \space mole \space AlCl_3 }$$
$$ 37.7 \space g \space Al \times \frac{1 \space mole \space Al }{ 26.98 \space g \space Al } \times \frac{ 2 \space moles \space AlCl_3 }{ 2 \space moles \space Al } \times \frac{ 133.33 \space g \space AlCl_3 }{1 \space mole \space AlCl_3 } = 186 \space g \space AlCl_3 $$
c. - Calculate or find the molar mass for $ Al $:
$ Al $ : 26.98 g/mol
- Using the molar mass as a conversion factor, find the amount in moles:
$$ 13.5 \space g \times \frac{1 \space mole}{ 26.98 \space g} = 0.500 \space mole$$
- 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:
$$ 8.00 \space g \times \frac{1 \space mole}{ 70.90 \space g} = 0.113 \space mole$$
Find the amount of product if each reactant is completely consumed.
$$ 0.500 \space mole \space Al \times \frac{ 2 \space moles \ AlCl_3 }{ 2 \space moles \space Al } = 0.500 \space mole \space AlCl_3 $$
$$ 0.113 \space mole \space Cl_2 \times \frac{ 2 \space moles \ AlCl_3 }{ 3 \space moles \space Cl_2 } = 0.0753 \space mole \space AlCl_3 $$
Since the reaction of $ Cl_2 $ produces less $ AlCl_3 $ for these quantities, it is the limiting reactant.
- Calculate or find the molar mass for $ AlCl_3 $:
$ AlCl_3 $ : ( 26.98 $\times$ 1 )+ ( 35.45 $\times$ 3 )= 133.33 g/mol
- Using the molar mass as a conversion factor, find the mass in g:
$$ 0.0753 \space mole \times \frac{ 133.33 \space g}{1 \space mole} = 10.0 \space g$$
d. - Calculate the theoretical yield:
- Calculate or find the molar mass for $ Al $:
$ Al $ : 26.98 g/mol
- Using the molar mass as a conversion factor, find the amount in moles:
$$ 45.0 \space g \times \frac{1 \space mole}{ 26.98 \space g} = 1.67 \space moles$$
- 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:
$$ 62.0 \space g \times \frac{1 \space mole}{ 70.90 \space g} = 0.874 \space mole$$
Find the amount of product if each reactant is completely consumed.
$$ 1.67 \space moles \space Al \times \frac{ 2 \space moles \ AlCl_3 }{ 2 \space moles \space Al } = 1.67 \space moles \space AlCl_3 $$
$$ 0.874 \space mole \space Cl_2 \times \frac{ 2 \space moles \ AlCl_3 }{ 3 \space moles \space Cl_2 } = 0.583 \space mole \space AlCl_3 $$
Since the reaction of $ Cl_2 $ produces less $ AlCl_3 $ for these quantities, it is the limiting reactant.
- Calculate or find the molar mass for $ AlCl_3 $:
$ AlCl_3 $ : 133.33 g/mol
- Using the molar mass as a conversion factor, find the mass in g:
$$ 0.583 \space mole \times \frac{ 133.33 \space g}{1 \space mole} = 77.7 \space g$$
- Calculate the percent yield:
$$\frac{66.5 \space g}{77.7 \space g} \times 100\% = 85.6\% $$
