Answer
a. The limiting reactant is $Cl_2$, and this reaction will produce $25.1$ g of $AlCl_3$.
b. The limiting reactant is $O_2$, and this reaction will produce $13.5$ g of $H_2O$.
c. The limiting reactant is $O_2$, and this reaction will produce $26.7$ g of $SO_2$.
Work Step by Step
- Calculate or find the molar mass for the compound:
$ Al $ : 26.98 g/mol
- Using the molar mass as a conversion factor, find the amount in moles:
$$ 20.0 \space g \times \frac{1 \space mole}{ 26.98 \space g} = 0.741 \space mole$$
- Calculate or find the molar mass for the compound:
$ Cl_2 $ : ( 35.45 $\times$ 2 )= 70.90 g/mol
- Using the molar mass as a conversion factor, find the amount in moles:
$$ 20.0 \space g \times \frac{1 \space mole}{ 70.90 \space g} = 0.282 \space mole$$
Find the amount of product if each reactant is completely consumed.
$$ 0.741 \space mole \space Al \times \frac{ 2 \space moles \ AlCl_3 }{ 2 \space moles \space Al } = 0.741 \space mole \space AlCl_3 $$
$$ 0.282 \space mole \space Cl_2 \times \frac{ 2 \space moles \ AlCl_3 }{ 3 \space moles \space Cl_2 } = 0.188 \space mole \space AlCl_3 $$
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 the compound:
$ 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:
$$ 0.188 \space mole \times \frac{ 133.33 \space g}{1 \space mole} = 25.1 \space g$$
------
b. - Calculate or find the molar mass for the compound:
$ NH_3 $ : ( 1.008 $\times$ 3 )+ ( 14.01 $\times$ 1 )= 17.03 g/mol
- Using the molar mass as a conversion factor, find the amount in moles:
$$ 20.0 \space g \times \frac{1 \space mole}{ 17.03 \space g} = 1.17 \space mole$$
- Calculate or find the molar mass for the compound:
$ O_2 $ : ( 16.00 $\times$ 2 )= 32.00 g/mol
- Using the molar mass as a conversion factor, find the amount in moles:
$$ 20.0 \space g \times \frac{1 \space mole}{ 32.00 \space g} = 0.625 \space mole$$
Find the amount of product if each reactant is completely consumed.
$$ 1.17 \space moles \space NH_3 \times \frac{ 6 \space moles \ H_2O }{ 4 \space moles \space NH_3 } = 1.75 \space moles \space H_2O $$
$$ 0.625 \space mole \space O_2 \times \frac{ 6 \space moles \ H_2O }{ 5 \space moles \space O_2 } = 0.750 \space mole \space H_2O $$
Since the reaction of $ O_2 $ produces less $ H_2O $ for these quantities, it is the limiting reactant.
- Calculate or find the molar mass for the compound:
$ H_2O $ : ( 1.008 $\times$ 2 )+ ( 16.00 $\times$ 1 )= 18.02 g/mol
- Using the molar mass as a conversion factor, find the mass in g:
$$ 0.750 \space mole \times \frac{ 18.02 \space g}{1 \space mole} = 13.5 \space g$$
------
- Calculate or find the molar mass for the compound:
$ CS_2 $ : ( 12.01 $\times$ 1 )+ ( 32.06 $\times$ 2 )= 76.13 g/mol
- Using the molar mass as a conversion factor, find the amount in moles:
$$ 20.0 \space g \times \frac{1 \space mole}{ 76.13 \space g} = 0.263 \space mole$$
- Calculate or find the molar mass for the compound:
$ O_2 $ : ( 16.00 $\times$ 2 )= 32.00 g/mol
- Using the molar mass as a conversion factor, find the amount in moles:
$$ 20.0 \space g \times \frac{1 \space mole}{ 32.00 \space g} = 0.625 \space mole$$
Find the amount of product if each reactant is completely consumed.
$$ \frac{ 1 \space mole \space CS_2 }{ 2 \space moles \space SO_2 } \space and \space \frac{ 2 \space moles \space SO_2 }{ 1 \space mole \space CS_2 }$$
$$ 0.263 \space mole \space CS_2 \times \frac{ 2 \space moles \ SO_2 }{ 1 \space mole \space CS_2 } = 0.526 \space mole \space SO_2 $$
$$ \frac{ 3 \space moles \space O_2 }{ 2 \space moles \space SO_2 } \space and \space \frac{ 2 \space moles \space SO_2 }{ 3 \space moles \space O_2 }$$
$$ 0.625 \space mole \space O_2 \times \frac{ 2 \space moles \ SO_2 }{ 3 \space moles \space O_2 } = 0.417 \space mole \space SO_2 $$
Since the reaction of $ O_2 $ produces less $ SO_2 $ for these quantities, it is the limiting reactant.
- Calculate or find the molar mass for the compound:
$ SO_2 $ : ( 16.00 $\times$ 2 )+ ( 32.06 $\times$ 1 )= 64.06 g/mol
- Using the molar mass as a conversion factor, find the mass in g:
$$ 0.417 \space mole \times \frac{ 64.06 \space g}{1 \space mole} = 26.7 \space g$$
