Answer
Theoretical yield: 20.6 g
Percent yield: 29.0 %
Work Step by Step
- Calculate or find the molar mass for $ Li $:
$ Li $ : 6.941 g/mol
- Using the molar mass as a conversion factor, find the amount in moles:
$$ 12.3 \space g \times \frac{1 \space mole}{ 6.941 \space g} = 1.77 \space moles$$
- Calculate or find the molar mass for $ N_2 $:
$ N_2 $ : ( 14.01 $\times$ 2 )= 28.02 g/mol
- Using the molar mass as a conversion factor, find the amount in moles:
$$ 33.6 \space g \times \frac{1 \space mole}{ 28.02 \space g} = 1.20 \space moles$$
Find the amount of product if each reactant is completely consumed.
$$ 1.77 \space moles \space Li \times \frac{ 2 \space moles \ Li_3N }{ 6 \space moles \space Li } = 0.590 \space mole \space Li_3N $$
$$ 1.20 \space moles \space N_2 \times \frac{ 2 \space moles \ Li_3N }{ 1 \space mole \space N_2 } = 2.40 \space moles \space Li_3N $$
Since the reaction of $ Li $ produces less $ Li_3N $ for these quantities, it is the limiting reactant.
- Calculate or find the molar mass for $ Li_3N $:
$ Li_3N $ : ( 6.941 $\times$ 3 )+ ( 14.01 $\times$ 1 )= 34.83 g/mol
- Using the molar mass as a conversion factor, find the mass in g:
$$ 0.590 \space mole \times \frac{ 34.83 \space g}{1 \space mole} = 20.6 \space g$$
- Calculate the percent yield
$$percent \space yield = \frac{5.89 \space g}{20.6 \space g} \times 100\% = 29.0 \%$$
