Question

1. NaOH mass of a solution of 200g in which its percentage is 25%. What mass of sulfuric acid solution is needed to completely neutralize the alkali. What mass of salt will be formed?

Answer 1

Answer:

$m_{H2SO4$ = 61.25 g

$m_{Na2SO4}$ = 88.75 g

Explanation:

$m_{NaOH}$ = $\frac{200 . 25 }{100}$ = 50 g

⇒ $n_{NaOH}$ = $\frac{50}{40}$ = 1.25 (moles)

2NaOH + H2SO4 ⇒ Na2SO4 + 2H2O

   2        :     1           :      1         :    2

 1.25                                                       (moles)

⇒  $n_{H2SO4}$ = 1.25 × 1 ÷ 2 = 0.625 (moles) ⇒ $m_{H2SO4}$ = 0.625 × 98 = 61.25 g

    $n_{Na2SO4}$ = 1.25 × 1 ÷ 2 = 0.625 (moles) ⇒$m_{Na2SO4}$ = 0.625 × 142 = 88.75 g