Question

$A+2B+3C\rightleftharpoons A{B}_2{C}_3$
Reaction of $6.0g$ of $A,6.0\times 10^{23}$ atoms of $B$, and $0.036mol$ of $C$ yields $4.8g$ of compound $A{B}_2{C}_3$. If the atomic mass of $A$ and $C$ are 60 and 80 amu respectively, the atomic mass of $B$ is (Avogadro no. $=6\times 10^{23}$ ):

• A. 70 amu
• B. 60 amu
• C. 50 amu
• D. 40 amu

Answer 1

Answer: (C)

Atomic mass of $A=60$ and atomic mass of $C=80$



Here $C$ is the limiting reagent, hence, $0.036$ mol of $C$ will completely consume $\frac{0.036}{3}=0.012mol$ of $A{B}_2{C}_3$ and will form $4.8gmmol$ of

$\left(A{B}_2{C}_3\right)=\frac{\text{ mass }}{\text{ mol. mass }}$

$0.012=\frac{4.8}{60+M_B\times 2+80\times 3}$

$\Rightarrow 300+2M_B=\frac{4.8}{(0.012)}=400$

$\Rightarrow 2M_B=100$

$\Rightarrow M_B=50amu$