Answer
a. 3.38 moles of $O_2$
b. 36.2 g of $Cr_2O_3$
c. 25.3 g of $Cr_2O_3$
d. Percent yield = 80.8 %
Work Step by Step
a. $$ 4.50 \space moles \space Cr \times \frac{ 3 \space moles \ O_2 }{ 4 \space moles \space Cr } = 3.38 \space moles \space O_2 $$
b. $ Cr $ : 52.00 g/mol
$$ \frac{1 \space mole \space Cr }{ 52.00 \space g \space Cr } \space and \space \frac{ 52.00 \space g \space Cr }{1 \space mole \space Cr }$$
$ Cr_2O_3 $ : ( 52.00 $\times$ 2 )+ ( 16.00 $\times$ 3 )= 152.00 g/mol
$$ \frac{1 \space mole \space Cr_2O_3 }{ 152.00 \space g \space Cr_2O_3 } \space and \space \frac{ 152.00 \space g \space Cr_2O_3 }{1 \space mole \space Cr_2O_3 }$$
$$ 24.8 \space g \space Cr \times \frac{1 \space mole \space Cr }{ 52.00 \space g \space Cr } \times \frac{ 2 \space moles \space Cr_2O_3 }{ 4 \space moles \space Cr } \times \frac{ 152.00 \space g \space Cr_2O_3 }{1 \space mole \space Cr_2O_3 } = 36.2 \space g \space Cr_2O_3 $$
c. - Calculate or find the molar mass for $ Cr $:
$ Cr $ : 52.00 g/mol
- Using the molar mass as a conversion factor, find the amount in moles:
$$ 26.0 \space g \times \frac{1 \space mole}{ 52.00 \space g} = 0.500 \space mole$$
- Calculate or find the molar mass for $ O_2 $:
$ O_2 $ : ( 16.00 $\times$ 2 )= 32.00 g/mol
- Using the molar mass as a conversion factor, find the amount in moles:
$$ 8.00 \space g \times \frac{1 \space mole}{ 32.00 \space g} = 0.250 \space mole$$
Find the amount of product if each reactant is completely consumed.
$$ 0.500 \space mole \space Cr \times \frac{ 2 \space moles \ Cr_2O_3 }{ 4 \space moles \space Cr } = 0.250 \space mole \space Cr_2O_3 $$
$$ 0.250 \space mole \space O_2 \times \frac{ 2 \space moles \ Cr_2O_3 }{ 3 \space moles \space O_2 } = 0.1667 \space mole \space Cr_2O_3 $$
Since the reaction of $ O_2 $ produces less $ Cr_2O_3 $ for these quantities, it is the limiting reactant.
- Calculate or find the molar mass for $ Cr_2O_3 $:
$ Cr_2O_3 $ : 152.00 g/mol
- Using the molar mass as a conversion factor, find the mass in g:
$$ 0.1667 \space mole \times \frac{ 152.00 \space g}{1 \space mole} = 25.3 \space g$$
d. - Find the theoretical yield:
- Calculate or find the molar mass for $ Cr $:
$ Cr $ : 52.00 g/mol
- Using the molar mass as a conversion factor, find the amount in moles:
$$ 74.0 \space g \times \frac{1 \space mole}{ 52.00 \space g} = 1.42 \space moles$$
- Calculate or find the molar mass for $ O_2 $:
$ O_2 $ : ( 16.00 $\times$ 2 )= 32.00 g/mol
- Using the molar mass as a conversion factor, find the amount in moles:
$$ 62.0 \space g \times \frac{1 \space mole}{ 32.00 \space g} = 1.94 \space moles$$
Find the amount of product if each reactant is completely consumed.
$$ 1.42 \space moles \space Cr \times \frac{ 2 \space moles \ Cr_2O_3 }{ 4 \space moles \space Cr } = 0.710 \space mole \space Cr_2O_3 $$
$$ 1.94 \space moles \space O_2 \times \frac{ 2 \space moles \ Cr_2O_3 }{ 3 \space moles \space O_2 } = 1.29 \space moles \space Cr_2O_3 $$
Since the reaction of $ Cr $ produces less $ Cr_2O_3 $ for these quantities, it is the limiting reactant.
- Calculate or find the molar mass for $ Cr_2O_3 $:
$ Cr_2O_3 $ : ( 52.00 $\times$ 2 )+ ( 16.00 $\times$ 3 )= 152.00 g/mol
- Using the molar mass as a conversion factor, find the mass in g:
$$ 0.710 \space mole \times \frac{ 152.00 \space g}{1 \space mole} = 108 \space g$$
- Calculate the percent yield:
$$\frac{87.3 \space g \space Cr_2O_3}{108 \space g \space Cr_2O_3} \times 100\% = 80.8 \%$$
