Question

The poisonous gas hydrogen cyanide (CN) is produced by the high-temperature reaction of ammonia with methane $\left(C{H}_4\right)$ hydrogen is also produced in this reaction.

Short answer

We are given the poisonous gas hydrogen cyanide (HCN) which is produced by the high-temperature reaction of ammonia with methane$\left(C{H}_4\right)$ , and the amount of methane mixed with ammonia.

Step-by-step solution

Step (1): Finding balanced equation for reaction

The unbalanced equation for the reaction of ammonia with methane is

$N{H}_3+C{H}_4\to HCN+H_2$

There are 7 moles of H on the left side. Hence, a coefficient of 3 must be there for hydrogen gas to balance the H atoms on both sides. The numbers of C and N are the same on both sides. Therefore, the balanced equation is

$N{H}_3+C{H}_4\to HCN+3H_2$

Step (2): Determination of molar mass of HCN,CH4,H2&NH3 

The following information is known:

Mass of methane= 500.0g

Mass of ammonia = 200.0g

We know the molar mass of N, C, H are 14.007g/mol, 12.011 g/mol and 1.0079 g/mol respectively. Then determine the molar masses of $HCN,C{H}_4,H_2\&N{H}_3$are follows,

$$\begin{gathered} ThemolarmassHCN=\left(1.0079+12.011+14.007\right)g/mol \\ =27.026g/mol \end{gathered}$$

role="math" localid="1663411973074" $$\begin{gathered} Themolarmassof\left(C{H}_4\right)=\left(12.011+4\times 1.0079\right)g/mol \\ =16.043g/mol \end{gathered}$$

$$\begin{gathered} Molarmassof{H}_2=\left(2\times 1.0079\right)g/mol \\ =2.016g/mol \end{gathered}$$

$$\begin{gathered} MolarmassofN{H}_3=\left(14.007+3\times 1.0079\right)g/mol \\ =12.031g/mol \end{gathered}$$

Step (3): Calculation of number of moles of NH3  and CH4  in 200g and 500g

We know that 1 mole of $N{H}_3$contains 17.03g.

Then, 1 g of $N{H}_3$ contains$\frac{1}{17.031}mol$ (or)

200g of $N{H}_3$ contains $=\frac{200}{17.031}mol=11.74mol$

Also, 1 mole of $\left(C{H}_4\right)$ contains 16.043g (or)

1g of $\left(C{H}_4\right)$contains $=\frac{1}{16.043g}mole$

500g of $\left(C{H}_4\right)$ contains =$\frac{500g}{16.043}mole$

= 31.17mol

Step (4): - Calculation of number of moles of HCN and H2  in balanced equation 

The reaction's balanced equation is

$N{H}_3+C{H}_4\to HCN+3H_2$

From the equation, 1 mole of $N{H}_3$ reacts with 1 mole of $\left(C{H}_4\right)$ to give 1 mole HCN and 3 moles of $H_2$ respectively. Hence after completion of reaction, the number of moles of HCN present is 11.74mol and the number of moles of $H_2$ is given by

(31.17-11.74) mol

=19.43 mol

Step (5): - Calculation of masses of CH4, HCN,  H2

By having these values, we can calculate the masses of $\left(C{H}_4\right)$ , HCN and $H_2$. These are as follows:

Mass of $\left(C{H}_4\right)$ present is given by

$$\begin{gathered} =\left(19.43\times 16.043\right)g \\ =311.7g \end{gathered}$$

The mass of$HCN=\left(11.74\times 27.026\right)g=312.3g$

Similarly, the mass of $H_2$ is $\left(3\times 11.74\times 2.016\right)g=71.00g$

Therefore, the masses of $\left(C{H}_4\right)$, HCN and $H_2$ are 311.7g, 317.3g and 71.00grespectively.