Answer:
Approximately [tex]1.195 \times 10^3 \; \rm g[/tex].
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:
[tex]M(\mathrm{NH_2}) = 14.007 + 2 \times 1.008 = 16.023\; \rm g \cdot mol^{-1}[/tex].
[tex]M(\mathrm{CO}_2) = 12.011 + 2\times 15.999 = 44.009\; \rm g \cdot mol^{-1}[/tex].
[tex]\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}[/tex]
Calculate the number of moles of each reactant:
[tex]\displaystyle n(\mathrm{NH}_2) = \frac{m}{M} \approx 39.7678\; \rm mol[/tex].
[tex]\displaystyle n(\mathrm{CO}_2) = \frac{m}{M} \approx 25.9492\; \rm mol[/tex].
One of the two reactants might be in excess. To find the limiting reactant, calculate the ratio [tex]\displaystyle \frac{n(\mathrm{NH}_2)}{n(\mathrm{CO}_2)}[/tex].
[tex]\displaystyle \frac{n(\mathrm{NH}_2)}{n(\mathrm{CO}_2)} \approx \frac{39.7678}{25.9492} \approx 1.53[/tex].
However, in the equation, the ratio between the coefficient of [tex]\mathrm{NH_2}[/tex] and [tex]\mathrm{CO_2}[/tex] is [tex]\displaystyle \frac{2}{1} = 2[/tex].
Since [tex]\displaystyle \frac{n(\mathrm{NH}_2)}{n(\mathrm{CO}_2)} \approx 1.53 < 2[/tex], the species on the numerator ([tex]\mathrm{NH_2}[/tex] in this case) is the limiting reactant.
As a result, [tex]n(\mathrm{(NH_2)_2CO})[/tex] should be calculated from [tex]n(\mathrm{NH_2})[/tex], not [tex]n(\mathrm{CO_2})[/tex].
In the equation, the coefficient ratio [tex]\displaystyle \frac{n(\mathrm{(NH_2)_2CO})}{n(\mathrm{NH_2})} = \frac{1}{2}[/tex].
Apply this ratio:
[tex]\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}[/tex].
Calculate [tex]m(\mathrm{(NH_2)_2CO})[/tex] using its molar mass value:
[tex]\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}[/tex].
Note: the last digit might be inaccurate due to rounding errors.
