Question

If 2 moles of NaOH reacts with 4 moles of HCl to give sodium chloride and then the produced sodium chloride reacts with 3 moles of concentrated sulphuric acid to give sodium sulphate, find the total moles of sodium sulphate produced. The concerned reactions are
$NaOH+HCl\left(diluted\right)\to NaCl+H_{2}O$
$2NaCl+H_{2}S{O}_4\to N{a}_{2}S{O}_4+2HCl$

• A. 2 mol
• B. 1.0 mol
• C. 4.2 mol
• D. 1.5 mol

Answer 1

The correct option is B 1.0 mol

$NaOH+HCl\left(diluted\right)\to NaCl+H_{2}O2NaCl+H_{2}S{O}_4\to N{a}_{2}S{O}_4+2HCl$

In reaction (i),
Limiting reagent will be NaOH since it is present in a smaller amount.
1 mol of NaOH produces 1 mol of NaCl
2 mol will produce 2 mol of NaCl
In reaction(ii),
2 mol of NaCl will produce 1 mol of $N{a}_{2}S{O}_4$.