Question

A mixture of 0.437 g of BaCl₂ • 2 H₂O (molar mass 244.27 g/mol) and 0.284 g Na₂SO₄ (molar mass 142.04 g/mol) is added to water. What is the theoretical yield of BaSO₄ precipitate (molar mass 233.39 g/mol)?

BaCl₂ • 2 H₂O + Na₂SO₄ --> BaSO₄ + 2 NaCl + 2 H₂O

Answer 1

BaCl2 • 2 H2O + Na2SO4 --> BaSO4 + 2 NaCl + 2 H2O this is a balanced equation

from this it is clear that one mole of BaSO4 will produce from the one mole of BaCl2 • 2 H2O and Na2SO4

no of moles of BaCl2 • 2 H2O = 0.437 g / 244.27 g/mol = 0.002 moles

no of moles of Na2SO4 = 0.284 g / 142.04 g/mol = 0.002 moles

almost both are 1:1 ratio

no of moles of BaSO4 formed = no of moles of either BaCl2 • 2 H2O or Na2SO4

moles of BaSO4 = 0.002 moles

theritical yield of BaSO4 in weight = no of moles of BaSO4 x molar mass of BaSO4   

= 0.002 moles x 233.39 g /mol

= 0.46678 grams of BaSo4 will form