Answer
a. 2.5 g of $AlCl_3$
b. 31.1 g of $AlCl_3$
c. 1.16 g of $AlCl_3$
Work Step by Step
a. - 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:
$$ 2.0 \space g \times \frac{1 \space mole}{ 26.98 \space g} = 0.074 \space mole$$
- 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:
$$ 2.0 \space g \times \frac{1 \space mole}{ 70.90 \space g} = 0.028 \space mole$$
Find the amount of product if each reactant is completely consumed.
$$ 0.074 \space mole \space Al \times \frac{ 2 \space moles \ AlCl_3 }{ 2 \space moles \space Al } = 0.074 \space mole \space AlCl_3 $$
$$ 0.028 \space mole \space Cl_2 \times \frac{ 2 \space moles \ AlCl_3 }{ 3 \space moles \space Cl_2 } = 0.019 \space mole \space AlCl_3 $$
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:
$$ 0.019 \space mole \times \frac{ 133.33 \space g}{1 \space mole} = 2.5 \space g$$
b.
- Using the molar mass as a conversion factor, find the amount in moles:
$$ 7.5 \space g \times \frac{1 \space mole}{ 26.98 \space g} = 0.28 \space mole$$
- Using the molar mass as a conversion factor, find the amount in moles:
$$ 24.8 \space g \times \frac{1 \space mole}{ 70.90 \space g} = 0.350 \space mole$$
Find the amount of product if each reactant is completely consumed.
$$ 0.28 \space mole \space Al \times \frac{ 2 \space moles \ AlCl_3 }{ 2 \space moles \space Al } = 0.28 \space mole \space AlCl_3 $$
$$ 0.350 \space mole \space Cl_2 \times \frac{ 2 \space moles \ AlCl_3 }{ 3 \space moles \space Cl_2 } = 0.233 \space mole \space AlCl_3 $$
Since the reaction of $ Cl_2 $ produces less $ AlCl_3 $ for these quantities, it is the limiting reactant.
- Using the molar mass as a conversion factor, find the mass in g:
$$ 0.233 \space mole \times \frac{ 133.33 \space g}{1 \space mole} = 31.1 \space g$$
c.
- Using the molar mass as a conversion factor, find the amount in moles:
$$ 0.235 \space g \times \frac{1 \space mole}{ 26.98 \space g} = 0.00871 \space mole$$
- Using the molar mass as a conversion factor, find the amount in moles:
$$ 1.15 \space g \times \frac{1 \space mole}{ 70.90 \space g} = 0.0162 \space mole$$
Find the amount of product if each reactant is completely consumed.
$$ 0.00871 \space mole \space Al \times \frac{ 2 \space moles \ AlCl_3 }{ 2 \space moles \space Al } = 0.00871 \space mole \space AlCl_3 $$
$$ 0.0162 \space mole \space Cl_2 \times \frac{ 2 \space moles \ AlCl_3 }{ 3 \space moles \space Cl_2 } = 0.0108 \space mole \space AlCl_3 $$
Since the reaction of $ Al $ produces less $ AlCl_3 $ for these quantities, it is the limiting reactant.
- Using the molar mass as a conversion factor, find the mass in g:
$$ 0.00871 \space mole \times \frac{ 133.33 \space g}{1 \space mole} = 1.16 \space g$$
