Chemistry Question on Stoichiometry and Stoichiometric Calculations

Question

For neutralization of one mol of NaOH the mass of 70%H 2SO4 required is :

Answer 1

Solution

The neutralization of $NaOH$ by $H _{2} SO _{4}$ takes place as follows $H _{2} SO _{4}+2 NaOH \longrightarrow Na _{2} SO _{4}+ H _{2} O$

For complete neutralization Equivalents of acid = equivalents of base Equivalents of $NaOH =$ moles $\times$ acidity $=1 \times 1=1$ Equivalents of $H _{2} SO _{4}=\frac{x}{98} \times 2=\frac{x}{49}$ (Mol. mass of $H _{2} SO _{4}=98$ ) Putting the values $1 \times 1 =\frac{x}{49}$ $\Rightarrow x =49\, g$ but $H _{2} SO _{4}$ is $70 \%$ let $y g 70 \% H _{2} SO _{4}$ is required $\frac{70}{100} \times y =49$ $\Rightarrow y =70\, g$