Answer
Limiting reactant: $Mg$
Theoretical yield: 16.7 g
Percent yield: 71.3%
Work Step by Step
- Calculate or find the molar mass for $ Mg $:
$ Mg $ : 24.31 g/mol
- Using the molar mass as a conversion factor, find the amount in moles:
$$ 10.1 \space g \times \frac{1 \space mole}{ 24.31 \space g} = 0.415 \space mole$$
- 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:
$$ 10.5 \space g \times \frac{1 \space mole}{ 32.00 \space g} = 0.328 \space mole$$
Find the amount of product if each reactant is completely consumed.
$$ 0.415 \space mole \space Mg \times \frac{ 2 \space moles \ MgO }{ 2 \space moles \space Mg } = 0.415 \space mole \space MgO $$
$$ 0.328 \space mole \space O_2 \times \frac{ 2 \space moles \ MgO }{ 1 \space mole \space O_2 } = 0.656 \space mole \space MgO $$
Since the reaction of $ Mg $ produces less $ MgO $ for these quantities, it is the limiting reactant.
- Calculate or find the molar mass for $ MgO $:
$ MgO $ : ( 24.31 $\times$ 1 )+ ( 16.00 $\times$ 1 )= 40.31 g/mol
- Using the molar mass as a conversion factor, find the mass in g:
$$ 0.415 \space mole \times \frac{ 40.31 \space g}{1 \space mole} = 16.7 \space g$$
$$Percent \space yield = \frac{ 11.9 }{ 16.7 } \times 100\% = 71.3\% $$
