## Chemistry: Molecular Approach (4th Edition)

$NH_3$ is the limiting reactant; The theoretical yield of urea is 240.5 g The percent yield for the reaction is 70.01%
- 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: $$4.005 \space mole \times \frac{ 60.06 \space g}{1 \space mole} = 240.5\underline{403} \space g = 240.5 \space g$$ $$Percent \space yield = \frac{ 168.4 }{ 240.5\underline{403} } \times 100\% = 70.01 \%$$