Q. 7.60 - An Introduction to General, Organic, and Biological Chemistry

Question

Nitrogen gas reacts with hydrogen gas to produce ammonia.

$N_2$(g) + $3H_2$(g) $\to$$2N{H}_3$(g)

(a) If you have $3.64$ g of H, how many grams of $N{H}_3$ can be

produced? 

(b) How many grams of H, are needed to react with $2.80$ g

of N? 

(c) How many grams of $N{H}_3$ can be produced from $12.0$ g

of H?

Step-by-Step Solution

Answer

(a) $41.25 \mathrm{g}$ of $N{H}_3$ will formed.

(b) $1.2 \mathrm{g}$ of H is needed for $2.80 \mathrm{g}$ of N.

(c) $136 \mathrm{g}$ of $N{H}_3$ will formed.

Part (a) Step 1: Given Information

We are given the $3.64 \mathrm{g}$ of H and we have to find out that how many grams of $N{H}_3$ will be formed.

Part (a) Step 2: Explanation

As we are given the mass of hydrogen and we have to find out the grams oh ammonium will formed so it is mass to mass conversion.

Now, $1$mole of H=$1$g of H

which gives the mole factor, $\frac{1 mole of H }{1 g of H}$

$3$ moles of H=$2$ moles of $N{H}_3$

which gives,$\frac{2 mole of N{H}_3}{3 mole of H}$

and $1$ mole of $N{H}_3$=$17 \mathrm{g}$ of $N{H}_3$

which gives $\frac{17 g of N{H}_3}{1 mole of N{H}_3}$

$3.64\cancel{g}\times \frac{1 \cancel{mole of H}}{1 \cancel{g of H}}\times \frac{17g of N{H}_3}{1\cancel{mole of N{H}_3}}\times \frac{2 \cancel{mole of N{H}_3}}{3\cancel{mole of H}}$

on solving these mole factors we get,

mass of $N{H}_3$ is $41.25 \mathrm{g}$.

Part (b) Step 1: Given Information

We are given the $2.80 \mathrm{g}$ of N and we have to find out that how many grams of H is needed to react.

Part (b) Step 2: Explanation

First, find out the number of moles given,

$2.80 \mathrm{g}$ of N=$0.4$mole of N

and $1$ mole of N=$3$moles of H

which gives, $\frac{3 moles H}{1 mole N}$

on solving, $0.4 \cancel{mole N}\times \frac{3 mole H}{1 \cancel{mole N}}$

We get, $1.2$ moles of H=$1.2$g of H.

Hence, $1.2$g of hydrogen is required.

Part (c) Step 1: Given Information

We are given the $12$g of H and we have to find out that how many grams of $N{H}_3$ will formed.

Part (b) Step 2: Explanation

First find out given number of moles,

$12 \mathrm{g}$ of H=$12$ moles of H

and $3$ moles of H= $2$ moles of $N{H}_3$,

which gives, $\frac{2 moles N{H}_3}{3 moles H}$

on solving, $12\cancel{ mole H}\times \frac{2 mole N{H}_3}{3 \cancel{mole H}}$

we get, $8$ mole of $N{H}_3$= 136 g of $N{H}_3$

Hence, $136 \mathrm{g}$ of $N{H}_3$ will formed.