Q. 7.64 - An Introduction to General, Organic, and Biological Chemistry

Question

Calcium cyanamide, $CaC{N}_2$ reacts with water to form calcium carbonate and ammonia.

$CaC{N}_2\left(s\right)+3H_{2}O\left(l\right)\to CaC{O}_3\left(s\right)+2N{H}_3\left(g\right)$

a. How many grams of $H_{2}O$ are needed to react with $75.0 \mathrm{g}$of $CaC{N}_2?$

b. How many grams of$N{H}_3$ are produced from $5.24 \mathrm{g}$ of $CaC{N}_2?$

c. How many grams of $CaC{O}_3$ form if  $155 \mathrm{g}$ of $H_{2}O$ reacts?

Step-by-Step Solution

Answer

(a) The $57.24 \mathrm{g}$ of $H_{2}O$ are needed to react with $75.0 \mathrm{g}$of $CaC{N}_2$.

(b) The $2.227 \mathrm{g}$ of $N{H}_3$are produced from $5.24 \mathrm{g}$of $CaC{N}_2$.

(c) The $287 \mathrm{g}$of $CaC{O}_3$ form if  $155 \mathrm{g}$of $H_{2}O$ reacts.

Part(a) Step 1: Given information

We have to find out the how many grams $H_{2}O$ are needed to react with $75.0 \mathrm{g}$of $CaC{N}_2$.

Part(a) Step 2: Explanation

Now, we solve this question,

$\frac{82 {\displaystyle \raisebox{1ex}{$\cancel{\mathrm{g}}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.}}{75\cancel{\mathrm{g}}}=1.06 \mathrm{moles}$

water moles that will react$=1.06\times 3 \mathrm{mole}$

$\underset{}{\therefore }$weight of water $=1.06\times 3\times 18 \mathrm{g}$

$=57.24 \mathrm{g}$.

Part(b) Step 1: Given information

We have to find out the how many grams of $N{H}_3$are produced from $5.24 \mathrm{g}$of $CaC{N}_2$.

Part(b) Step 2: Explanation

Now, we solve this question,

moles of ${\mathrm{CaCN}}_2=\frac{5.24 \cancel{\mathrm{g}}}{80 \cancel{\mathrm{g}}/\mathrm{mol}}=0.0655 \mathrm{mol}$,

moles of ${\mathrm{NH}}_3=2\times 0.0655 \mathrm{mol}$,

weight of ${\mathrm{NH}}_3=\left(2\times 0.0655\right) \cancel{\mathrm{mol} }\times 17 \frac{\mathrm{g}}{\cancel{\mathrm{mol}}}$,

$=2.227 \mathrm{g}$.

Part(c) Step 1: Given information

We have to find out the how many grams of $CaC{O}_3$from if $155 \mathrm{g}$of $H_{2}O$ reacts.

Part(c) Step 2: Explanation

Now, we solve this question,

$\frac{155 \cancel{\mathrm{g}}}{18 \cancel{\mathrm{g}}/\mathrm{mol}}=8.611 \mathrm{mol}$(moles of water given),

$3$moles of water $\to$$1$moles of $CaC{O}_3$,

$\therefore$$8.611$moles of water $=\frac{1 \mathrm{mol}\times 8.611 \cancel{\mathrm{mol}}}{3 \cancel{\mathrm{mol}}}=2.87$ moles of $CaC{O}_3$,

weight of $CaC{O}_3$$=2.87 \cancel{\mathrm{mol}} \times 100 \mathrm{g}/\cancel{\mathrm{mol}}$,

$=287 \mathrm{g}$.