Question

In a reaction 3A+ $B_2$ $⟶A_3{B}_2$ if 180 atoms of A and 100 molecules of B react then

1) $B_2$ is limiting reagent and 100 molecules of $A_3{B}_2$ will be formed.

2) A is limiting reagent and 60 molecules of $A_3{B}_2$ is formed

3) A is limiting reagent and 180 molecules of $A_3{B}_2$ will be formed

4) B is limiting reactant and 60 molecules of $A_3{B}_2$ will be formed.

Answer 1

Step 1:

The given reaction is:

$3A+B_2⟶A_3{B}_2$

180 atoms of A and 100 molecules of B.

Step 2:

We know that,

$n\left(moles\right)=\frac{N_0}{N_A}$ (molecules/atoms)

$n_A=\frac{180}{N_A}moles{n}_{B_2}=\frac{100}{N_A}moles$

Step 3: To find the limiting reagent, we need to divide no. of moles and coefficient.

For A, $\frac{180}{N_A}$/ 3

For B, $\frac{100}{N_A}$ / 1

Hence, the limiting reagent is A.

The reaction will proceed as per from the balanced reaction we can say 3 moles of A gives 1 mole of $A_3{B}_2$.

$\therefore$ 1 mole of A $⟶$ $\frac{1}{3}$ mole of A

$\therefore$ $\frac{180}{N_A}$mole of A will give $\frac{1}{3}\times \frac{180}{N_A}molesof{A}_3{B}_2=\frac{60}{N_A}$

$\therefore n=\frac{N_0}{N_A}$

Number of molecules = $\frac{60}{N_A}\times N_A=60moleculesof{A}_3{B}_2$

Hence, option (2) is correct.