Chemistry Question on Stoichiometry and Stoichiometric Calculations

Question

The mass of 70% $H_2SO_4$ by mass required for neutralisation of 1 mol of NaOH is

Answer 1

Solution

$2NaOH + H_{2}SO_{4} \to Na_{2}SO_{4}+H_{2}O$ Pure $H_{2}SO_{4}$ required for $1$ mole of $NaOH$ $=\frac{1}{2}$ mol $=49\,g$ $70\%$ $H_{2}SO_{4}$ required for $1$ mole of $NaOH$ $=\frac{49\times100}{70}$ $=70\,g$