Q. 7.61 - An Introduction to General, Organic, and Biological Chemistry

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 with$13.6 \mathrm{g}$ of $N{H}_3?$

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

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

Step-by-Step Solution

Answer

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

(b) The $1.89$ g of $N_2$ can be produced when $6.50$ g of $O_2$reacts.

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

Part(a) Step 1: Given information

We have to find out how many grams of $O_2$are needed to react with $13.6 \mathrm{g}$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 when $6.50 \mathrm{g}$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$,

$1 \mathrm{g} {\mathrm{O}}_2$produces $\frac{28 \cancel{\mathrm{g}}}{96 \cancel{\mathrm{g}}} {\mathrm{N}}_2$,

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

$=1.89 \mathrm{g} {\mathrm{N}}_2$.

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

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