Question

For the reaction:
$A\left(g\right)+B\left(g\right)\to C\left(g\right)$
At STP 30 g of gas $A$ reacts with 11.2 L of gas B and produces gas C, which is diatomic. Calculate the number of atoms in C.
(Molar mass of A = 60 g/mol, $N_A$ is Avogadro's number )

• A. $N_A$
• B. $0.05N_A$
• C. $0.5N_A$
• D. $1.5N_A$

Answer 1

The correct option is A $N_A$

For the reaction:
$A\left(g\right)+B\left(g\right)\to C\left(g\right)$
Number of moles of A(g)
$n_A=\frac{givenmass}{molarmass}=\frac{30}{60}=0.5mol$
Number of moles of B(g)
$n_B=\frac{\text{Volume(L) of gas at STP}}{22.4L}=\frac{11.2}{22.4}=0.5mol$

0.5 mol of each A and B will produce 0.5 mol of C.
$\underset{0.5mol}{A\left(g\right)}+\underset{0.5mol}{B\left(g\right)}\to \underset{0.5mol}{C\left(g\right)}$
We know that:
Number of moles of C gas $=n_C=0.5$
$\text{Number of atoms of C}=n_C\times N_A\times 2=N_A$