Question

how many grams of barium sulfate are produced if 25.34 mL of 0.113M BaCl2 completely react given the reaction: BaCl2(aq)+Na2SO4(aq)=BaSO4(s)+2NaCl(aq)

Answer 1

Consider reaction , BaCl₂ (aq) + Na₂SO₄ (aq)  $\rightarrow$ BaSO₄ (s)+2 NaCl (aq)

From reaction, 1 mol  BaCl₂ $\equiv$ 1 mol Na₂SO₄$\equiv$ 1 mol BaSO₄  $\equiv$ 2 mol NaCl

i e 1 mole  BaSO₄ will be produced from 1 mole  BaCl₂.

We have, [  BaCl₂ ] = No. of moles of  BaCl₂ / volume of solution in L

$\therefore$ No. of moles of  BaCl₂ = [  BaCl₂ ] $\times$ volume of solution in L

Here we have,  [  BaCl₂ ] = 0.113 M and volume of solution = 25.34 ml = 0.02534 ml.

$\therefore$ No. of moles of  BaCl₂ = 0.113 mol / L $\times$ 0.02534 L = 0.002863 mol

We have relation, 1 mol  BaCl₂ $\equiv$ 1 mol BaSO₄

Hence, No. of moles of  BaCl₂ reacted = no. of moles of BaSO₄ produced = 0.002863 mol

We have , No. of moles = Mass / Molar mass

$\therefore$ Mass = No. of moles $\times$ Molar mass

Molar mass of BaSO₄ = 137.34 + 32.06 + ( 4 $\times$ 16.00) = 233.40 g/mol

$\therefore$ Mass BaSO₄ produced in the reaction = 0.002863 mol $\times$ 233.40 g /mol = 0.668 g

ANSWER :  Mass BaSO₄ produced in the reaction = 0.668 g