Question

HCI + Na2SO4 > NaCI + H2SO4

A. What is my theoretical yield of NaCI if I start with 300 g of Na2SO4?

B. If my percent yield was 67%, what was my actual yield?

Answer 1

The given reaction is mentioned below

$HCl + Na_2SO_4 \longrightarrow NaCl + H_2SO_4$

we will write balanced chemical equation for it

$2HCl + Na_2SO_4 \longrightarrow 2NaCl + H_2SO_4$

here , one mole of $Na_2SO_4$ will give 2 mole of NaCl

gram molecular mass of Na₂SO₄ is equal to = 142 g/mole

gram molecular mass of NaCl = 58.5 g/mole

A.

here , given mass of Na₂SO₄ is 300 g

so mole of $Na_2SO_4 = \frac{given mass}{molar mass} = \frac {300}{142} = 2.113$

as per above balanced equation , 1 mole of Na₂SO₄ give 2 mole of HCl

so 2.113 mole of Na₂SO₄ will give $= 2\times 2.113$ mole of NaCl

so, mass of NaCl produced=$2\times2.113\times58.5=247.22 g$ of NaCl

This is the theoretical yield in above case

B.

As it is given that percent yield was 67 %

so let x be the actual yield

now as per question

$67$ % of x = 247.22 g

so , x$=247.22 \times \frac{100}{67}$ = 368.98 g

so,actual theoretical yield was = 368.98 g of NaCl