Question

Ammonia and oxygen react to form nitrogen and water.

$4N{H}_3\left(g\right)+3O_2\left(g\right)\to 2N_2\left(g\right)+6H_{2}O\left(g\right)$

a. How many grams of $O_2$are needed to react withlocalid="1653487157133" $13.6\mathrm{g}$of $N{H}_3?$

b. How many grams of $N_2$can be produced whenlocalid="1653487174328" $6.50\mathrm{g}$of $O_2$reacts?

c. How many grams of $H_{2}O$are formed from the reaction of localid="1653487189643" $34.0\mathrm{g}$of$N{H}_3?$

Short answer

(a) The $19.2$g of $O_2$are needed to react with localid="1653487404761" $13.6$g of $N{H}_3$.

(b) The localid="1653487292397" $1.89$g of $N_2$can be produced when localid="1653487421465" $6.50$g of $O_2$reacts.

(c) The localid="1653487335142" $54$g of ${\mathrm{H}}_2\mathrm{O}$are formed from the reaction of$34.0$g of$N{H}_3$.

Step-by-step solution

Part(a) Step 1: Given information

We have to find out how many grams of$O_2$are needed to react withlocalid="1653487528811" $13.6\mathrm{g}$of$N{H}_3$.

Part(a) Step 2: Explanation

Now, we solve this question,

$68\mathrm{g}{\mathrm{NH}}_3$reacts with $96\mathrm{g}{\mathrm{O}}_2$,

$1\mathrm{g}{\mathrm{NH}}_3$reacts with $\frac{96\cancel{\mathrm{g}}}{68\cancel{\mathrm{g}}}{\mathrm{O}}_2$,

So,$13.6\mathrm{g}{\mathrm{NH}}_3$reacts with $\frac{96\cancel{\mathrm{g}}\times 13.6\mathrm{g}}{68\cancel{\mathrm{g}}}{\mathrm{O}}_2$,

$=19.2\mathrm{g}{\mathrm{O}}_2$.

Part(b) Step 1: Given information

We have to find out how many grams of$N_2$can be produced whenlocalid="1653487802710" $6.50\mathrm{g}$of$O_2$reacts.

Part(b) Step 2: Explanation

Now, we solve this question,

$96\mathrm{g}{\mathrm{O}}_2$produces $28\mathrm{g}$$N_2$,

localid="1653487899826" $1\mathrm{g}{\mathrm{O}}_2$produces localid="1653487936104" $\frac{28\cancel{\mathrm{g}}}{96\cancel{\mathrm{g}}}{\mathrm{N}}_2$,

localid="1653487971697" $6.5\mathrm{g}{\mathrm{O}}_2$produces localid="1653488017810" $=\frac{28\cancel{\mathrm{g}}\times 6.50\mathrm{g}}{96\cancel{\mathrm{g}}}{\mathrm{N}}_2$,

localid="1653488038297" $=1.89\mathrm{g}{\mathrm{N}}_2$.

Part(c) Step 1: Given information

We have to find out how many grams of ${\mathrm{H}}_2\mathrm{O}$are formed from the reaction of$34.0\mathrm{g}$of$N{H}_3$.

Part(c) Step 2: Explanation

Now, we solve this question,

$68\mathrm{g}{\mathrm{NH}}_3$forms $108\mathrm{g}{\mathrm{H}}_2\mathrm{O}$,

$1\mathrm{g}{\mathrm{NH}}_3$forms $\frac{108\cancel{\mathrm{g}}}{68\cancel{\mathrm{g}}}{\mathrm{H}}_2\mathrm{O}$,

$34.0\mathrm{g}{\mathrm{NH}}_3$forms $=\frac{108\cancel{\mathrm{g}}\times 34\mathrm{g}}{68\cancel{\mathrm{g}}}{\mathrm{H}}_2\mathrm{O}$,

$=54\mathrm{g}{\mathrm{H}}_2\mathrm{O}$.