Question

2NH2 + CO2 -> (NH2)2CO + H2O

Answer 1

Answer:

Approximately $1.195 \times 10^3 \; \rm g$.

Explanation:

Look the relative atomic mass of each element on a modern periodic table:

• H: 1.008.
• N: 14.007.
• O: 15.999.
• C: 12.011.

Calculate molar mass values from relative atomic mass:

$M(\mathrm{NH_2}) = 14.007 + 2 \times 1.008 = 16.023\; \rm g \cdot mol^{-1}$.

$M(\mathrm{CO}_2) = 12.011 + 2\times 15.999 = 44.009\; \rm g \cdot mol^{-1}$.

$\begin{aligned}&M(\mathrm{(NH_2)_2CO}) \\ &= 2 \times (14.007 + 2 \times 1.008) + 12.011 + 15.999 \\ &= 60.056\; \rm g \cdot mol^{-1}\end{aligned}$

Calculate the number of moles of each reactant:

$\displaystyle n(\mathrm{NH}_2) = \frac{m}{M} \approx 39.7678\; \rm mol$.

$\displaystyle n(\mathrm{CO}_2) = \frac{m}{M} \approx 25.9492\; \rm mol$.

One of the two reactants might be in excess. To find the limiting reactant, calculate the ratio $\displaystyle \frac{n(\mathrm{NH}_2)}{n(\mathrm{CO}_2)}$.

$\displaystyle \frac{n(\mathrm{NH}_2)}{n(\mathrm{CO}_2)} \approx \frac{39.7678}{25.9492} \approx 1.53$.

However, in the equation, the ratio between the coefficient of $\mathrm{NH_2}$ and $\mathrm{CO_2}$ is $\displaystyle \frac{2}{1} = 2$.

Since $\displaystyle \frac{n(\mathrm{NH}_2)}{n(\mathrm{CO}_2)} \approx 1.53 < 2$, the species on the numerator ($\mathrm{NH_2}$ in this case) is the limiting reactant.

As a result, $n(\mathrm{(NH_2)_2CO})$ should be calculated from $n(\mathrm{NH_2})$, not $n(\mathrm{CO_2})$.

In the equation, the coefficient ratio $\displaystyle \frac{n(\mathrm{(NH_2)_2CO})}{n(\mathrm{NH_2})} = \frac{1}{2}$.

Apply this ratio:

$\begin{aligned}& n(\mathrm{(NH_2)_2CO}) \\ &= \displaystyle \frac{n(\mathrm{(NH_2)_2CO})}{n(\mathrm{NH_2})} \cdot n(\mathrm{NH_2}) \\ &\approx \frac{1}{2} \times 39.7678 \\ & \approx 19.8839\; \rm mol \end{aligned}$.

Calculate $m(\mathrm{(NH_2)_2CO})$ using its molar mass value:

$\begin{aligned}&m(\mathrm{(NH_2)_2CO}) \\ &= n(\mathrm{(NH_2)_2CO}) \cdot M(\mathrm{(NH_2)_2CO})\\&\approx 19.8839 \times 60.056 \\ &\approx 1.195 \times 10^3\; \rm g \cdot mol^{-1}\end{aligned}$.

Note: the last digit might be inaccurate due to rounding errors.