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?

Short answer

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

(b) localid="1654092532964" $1.2\mathrm{g}$of H is needed for localid="1654092551826" $2.80\mathrm{g}$of N.

(c)localid="1654092541424" $136\mathrm{g}$of$N{H}_3$will formed.

Step-by-step solution

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, $\frac{1moleofH}{1gofH}$

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

which gives,localid="1652502982783" $\frac{2moleofN{H}_3}{3moleofH}$

and $1$mole of $N{H}_3$=localid="1654092582751" $17\mathrm{g}$of $N{H}_3$

which gives localid="1652502944715" $\frac{17gofN{H}_3}{1moleofN{H}_3}$

$3.64\cancel{g}\times \frac{1\cancel{moleofH}}{1\cancel{gofH}}\times \frac{17gofN{H}_3}{1\cancel{moleofN{H}_3}}\times \frac{2\cancel{moleofN{H}_3}}{3\cancel{moleofH}}$

on solving these mole factors we get,

mass of$N{H}_3$islocalid="1654092590029" $41.25\mathrm{g}$.

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, $\frac{3molesH}{1moleN}$

on solving, localid="1653488271984" $0.4\cancel{moleN}\times \frac{3moleH}{1\cancel{moleN}}$

We get, $1.2$moles of H=$1.2$g of H.

Hence, $1.2$gof 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, $\frac{2molesN{H}_3}{3molesH}$

on solving, localid="1653488309919" $12\cancel{moleH}\times \frac{2moleN{H}_3}{3\cancel{moleH}}$

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

Hence, localid="1654092636554" $136\mathrm{g}$of $N{H}_3$will formed.