Answer
$NH_3$ is the limiting reactant;
The theoretical yield of urea is 240.5 g
The percent yield for the reaction is 70.01%
Work Step by Step
- Calculate or find the molar mass for $ NH_3 $:
$ 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:
$$ 136.4 \space g \times \frac{1 \space mole}{ 17.03 \space g} = 8.009 \space moles$$
- Calculate or find the molar mass for $ CO_2 $:
$ CO_2 $ : ( 12.01 $\times$ 1 )+ ( 16.00 $\times$ 2 )= 44.01 g/mol
- Using the molar mass as a conversion factor, find the amount in moles:
$$ 211.4 \space g \times \frac{1 \space mole}{ 44.01 \space g} = 4.803 \space moles$$
Find the amount of product if each reactant is completely consumed.
$$ 8.009 \space moles \space NH_3 \times \frac{ 1 \space mole \ CH_4N_2O }{ 2 \space moles \space NH_3 } = 4.005 \space moles \space CH_4N_2O $$
$$ 4.803 \space moles \space CO_2 \times \frac{ 1 \space mole \ CH_4N_2O }{ 1 \space mole \space CO_2 } = 4.803 \space moles \space CH_4N_2O $$
Since the reaction of $ NH_3 $ produces less $ CH_4N_2O $ for these quantities, it is the limiting reactant.
- Calculate or find the molar mass for $ CH_4N_2O $:
$ CH_4N_2O $ : ( 1.008 $\times$ 4 )+ ( 12.01 $\times$ 1 )+ ( 14.01 $\times$ 2 )+ ( 16.00 $\times$ 1 )= 60.06 g/mol
- Using the molar mass as a conversion factor, find the mass in g:
$$ 4.005 \space mole \times \frac{ 60.06 \space g}{1 \space mole} = 240.5\underline{403} \space g = 240.5 \space g$$
$$Percent \space yield = \frac{ 168.4 }{ 240.5\underline{403} } \times 100\% = 70.01 \% $$
