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 localid="1653494390919" $75.0\mathrm{g}$of $CaC{N}_2?$

b. How many grams of$N{H}_3$are produced from localid="1653494416163" $5.24\mathrm{g}$of $CaC{N}_2?$

c. How many grams of $CaC{O}_3$form iflocalid="1653494442688" $155\mathrm{g}$of $H_{2}O$reacts?

Short answer

(a) The$57.24\mathrm{g}$of $H_{2}O$are needed to react with localid="1653494592001" $75.0\mathrm{g}$of $CaC{N}_2$.

(b) Thelocalid="1653494503814" $2.227\mathrm{g}$of $N{H}_3$are produced from localid="1653494554644" $5.24\mathrm{g}$of $CaC{N}_2$.

(c) Thelocalid="1653494518404" $287\mathrm{g}$of$CaC{O}_3$form iflocalid="1653494540479" $155\mathrm{g}$of$H_{2}O$reacts.

Step-by-step solution

Part(a) Step 1: Given information

We have to find out the how many grams $H_{2}O$are needed to react with localid="1653494626093" $75.0\mathrm{g}$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 waterlocalid="1653494807736" role="math" $=1.06\times 3\times 18\mathrm{g}$

localid="1653494829299" $=57.24\mathrm{g}$.

Part(b) Step 1: Given information

We have to find out the how many grams of $N{H}_3$are produced from localid="1653494766823" $5.24\mathrm{g}$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 localid="1653495090999" $155\mathrm{g}$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 localid="1653495197251" $=\frac{1\mathrm{mol}\times 8.611\cancel{\mathrm{mol}}}{3\cancel{\mathrm{mol}}}=2.87$moles of $CaC{O}_3$,

weight of $CaC{O}_3$localid="1653495237884" $=2.87\cancel{\mathrm{mol}}\times 100\mathrm{g}/\cancel{\mathrm{mol}}$,

localid="1653495258912" $=287\mathrm{g}$.