Answer
a. Limiting reactant: $Al$. 37.8 g of $Al_2O_3$.
b. Limiting reactant: $NO_2$. 18.3 g of $HNO_3$.
c. Limiting reactant: $O_2$. 11.3 g of $H_2O$.
Work Step by Step
a. - 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:
$ 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.
$$ 0.741 \space mole \space Al \times \frac{ 2 \space moles \ Al_2O_3 }{ 4 \space moles \space Al } = 0.371 \space mole \space Al_2O_3 $$
$$ 0.625 \space mole \space O_2 \times \frac{ 2 \space moles \ Al_2O_3 }{ 3 \space moles \space O_2 } = 0.417 \space mole \space Al_2O_3 $$
Since the reaction of $ Al $ produces less $ Al_2O_3 $ for these quantities, it is the limiting reactant.
- Calculate or find the molar mass for the compound:
$ Al_2O_3 $ : ( 26.98 $\times$ 2 )+ ( 16.00 $\times$ 3 )= 101.96 g/mol
- Using the molar mass as a conversion factor, find the mass in g:
$$ 0.371 \space mole \times \frac{ 101.96 \space g}{1 \space mole} = 37.8 \space g$$
b. - Calculate or find the molar mass for the compound:
$ NO_2 $ : ( 14.01 $\times$ 1 )+ ( 16.00 $\times$ 2 )= 46.01 g/mol
- Using the molar mass as a conversion factor, find the amount in moles:
$$ 20.0 \space g \times \frac{1 \space mole}{ 46.01 \space g} = 0.435 \space mole$$
- 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 amount in moles:
$$ 20.0 \space g \times \frac{1 \space mole}{ 18.02 \space g} = 1.11 \space mole$$
Find the amount of product if each reactant is completely consumed.
$$ 0.435 \space mole \space NO_2 \times \frac{ 2 \space moles \ HNO_3 }{ 3 \space moles \space NO_2 } = 0.290 \space mole \space HNO_3 $$
$$ 1.11 \space moles \space H_2O \times \frac{ 2 \space moles \ HNO_3 }{ 1 \space mole \space H_2O } = 2.22 \space moles \space HNO_3 $$
Since the reaction of $ NO_2 $ produces less $ HNO_3 $ for these quantities, it is the limiting reactant.
- Calculate or find the molar mass for the compound:
$ HNO_3 $ : ( 1.008 $\times$ 1 )+ ( 14.01 $\times$ 1 )+ ( 16.00 $\times$ 3 )= 63.02 g/mol
- Using the molar mass as a conversion factor, find the mass in g:
$$ 0.290 \space mole \times \frac{ 63.02 \space g}{1 \space mole} = 18.3 \space g$$
c.- Calculate or find the molar mass for the compound:
$ C_2H_5OH $ : ( 1.008 $\times$ 6 )+ ( 12.01 $\times$ 2 )+ ( 16.00 $\times$ 1 )= 46.07 g/mol
- Using the molar mass as a conversion factor, find the amount in moles:
$$ 20.0 \space g \times \frac{1 \space mole}{ 46.07 \space g} = 0.434 \space mole$$
- 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.
$$ 0.434 \space mole \space C_2H_5OH \times \frac{ 3 \space moles \ H_2O }{ 1 \space mole \space C_2H_5OH } = 1.30 \space moles \space H_2O $$
$$ 0.625 \space mole \space O_2 \times \frac{ 3 \space moles \ H_2O }{ 3 \space moles \space O_2 } = 0.625 \space mole \space H_2O $$
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:
$$ 0.625 \space mole \times \frac{ 18.02 \space g}{1 \space mole} = 11.3 \space g$$
