Question

How many molecules of acetylene would be formed when $10.0 g$ of calcium carbide is treated with water? (Avogardro's constant $=6.02 \times 10^{23} / mole ; Ca =40, C =12, O =16$ and $H =1$)

(A) $9.4 \times 10^{22}$

(B) $94 \times 10^{20}$

(C) $9.4 \times 10^{23}$

(D) $18.8 \times 10^{22}$

Answer 1

ANSWER : Here, the correct answer of this question will be option (A) which is — $9.4 \times 10^{22}$

Explanation :

$CaC _{2}+2 H _{2} O \rightarrow Ca ( OH )_{2}+ C _{2} H _{2}$

$\begin{aligned}40 &+12 \times 2 \\& 64 g \\\therefore & 10 g CaC C_{2}=\frac{26}{64} \times 10 g C _{2} H _{2}\end{aligned}$

$\therefore$ No of moles of $C _{2} H _{2}=\frac{26}{64} \times \frac{10}{26}$

$\therefore$ No of molecules of $C_{2} H_{2}=\frac{10}{64} \times 6.022 \times 10^{23}$

$=\frac{60.22}{64} \times 10^{23} \\=9.4 \times 10^{22}$