Question

acrylonitrile is an important building block for synthetic fibers and plastics. over 1.4 billion kilograms of acrylonitrile are produced in the united states each year. the compound is synthesized from propene (c₃h₆) in the following reaction: 2c₃h₆(g) + 2nh₃(g) + 3o₂(g) → 2c₃h₃n(l) + 6h₂o(g). determine how many kg of acrylonitrile can be prepared from 1490 kg of propene, 781 kg of ammonia, and 1980 kg of oxygen. kg
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Explanation:

Step1: Calculate molar masses

The molar mass of propene ($C_3H_6$) is $M_{C_3H_6}=3\times12.01 + 6\times1.01=42.09\ g/mol$. The molar mass of acrylonitrile ($C_3H_3N$) is $M_{C_3H_3N}=3\times12.01+3\times1.01 + 14.01=53.06\ g/mol$.

Step2: Determine the limiting - reactant

The balanced chemical equation is $2C_3H_6(g)+2NH_3(g)+3O_2(g)
ightarrow2C_3H_3N(l)+6H_2O(g)$.
The number of moles of propene $n_{C_3H_6}=\frac{m_{C_3H_6}}{M_{C_3H_6}}=\frac{1490\times10^{3}\ g}{42.09\ g/mol}\approx3.54\times10^{4}\ mol$.
The number of moles of ammonia $n_{NH_3}=\frac{m_{NH_3}}{M_{NH_3}}=\frac{781\times10^{3}\ g}{17.03\ g/mol}\approx4.59\times10^{4}\ mol$.
The number of moles of oxygen $n_{O_2}=\frac{m_{O_2}}{M_{O_2}}=\frac{1980\times10^{3}\ g}{32.00\ g/mol}\approx6.19\times10^{4}\ mol$.
From the stoichiometry of the reaction, the mole - ratio of $C_3H_6:NH_3:O_2 = 2:2:3$.
For propene and ammonia, the mole - ratio is $1:1$. Since $n_{C_3H_6}For propene and oxygen, the mole - ratio of $C_3H_6$ to $O_2$ is $\frac{n_{C_3H_6}}{n_{O_2}}=\frac{3.54\times10^{4}\ mol}{6.19\times10^{4}\ mol}\approx0.57$. According to the stoichiometry, the ratio should be $\frac{2}{3}\approx0.67$. So, oxygen is the limiting reactant.

Step3: Calculate the mass of acrylonitrile produced

From the stoichiometry, the mole - ratio of $O_2$ to $C_3H_3N$ is $3:2$.
The number of moles of acrylonitrile produced $n_{C_3H_3N}=\frac{2}{3}n_{O_2}=\frac{2}{3}\times6.19\times10^{4}\ mol\approx4.13\times10^{4}\ mol$.
The mass of acrylonitrile produced $m_{C_3H_3N}=n_{C_3H_3N}\times M_{C_3H_3N}=4.13\times10^{4}\ mol\times53.06\ g/mol\approx2.19\times10^{6}\ g = 2190\ kg$.

Answer:

2190