Question

Match the Column-X with Column-Y:
Given reaction: $A+B_2\to A{B}_2$
$\begin{array}{cc}\text{Column X}& \text{Column Y}\\ \text{(i) 300 atoms of A + 200 molecules of}{B}_2& \text{(p)}{B}_2\text{is limiting reagent}\\ \text{(ii) 100 atoms of A + 100 molecules of}{B}_2& \text{(q)}A\text{is limiting reagent}\\ \text{(iii) 5 mol of A + 2.5 mol of}{B}_2& \text{(r) None of the reactant is in excess}\\ \text{(iv) 2.5 mol of A + 5 mol of}{B}_2& \text{(s) 200 molecules of}A{B}_2\text{will be formed}\end{array}$

• A. (i- p, s); (ii- r); (iii- p); (iv- q)
• B. (i- s); (ii- r); (iii- q); (iv- p)
• C. (i- s); (ii- p); (iii- r); (iv- r)
• D. (i- p, s); (ii- r); (iii- q); (iv- p)

Answer 1

The correct option is A (i- p, s); (ii- r); (iii- p); (iv- q)

In option i),
As per the reaction, 1 atom of A reacts with 1 molecule of B which means 200 atoms of A will react with 200 molecules of B leaving behind 100 atoms of A. Hence, $B_2$ is the limiting reagent and 200 molecules of $A{B}_2$ will be formed.
In option ii),
100 atoms of A will react with 100 molecules of $B_2$ as per the stoichiometry.
None of the reactants will be left.
In option iii),
1 mole of A reacts with 1 mole of $B_2$ to form 1 mole of $A{B}_2$. Hence, 2.5 mole of $B_2$ will react will 2.5 mole of A leaving behind 2.5 moles of A. $B_2$ is the limiting reagent.
In option iv),
2.5 moles of A will react with 2.5 moles of $B_2$. Hence, A will be the limiting reagent.