Question

Magnesium sulfate reacts with barium chloride according to the following balanced equation. If 1.20 g of magnesium sulfate is allowed to react with 10.0 g of barium chloride in a water solution, what is the theoretical yield of barium sulfate?

MgSO₄(aq) + BaCl₂(aq) → BaSO₄(s) + MgCl₂(aq)

Answer 1

MgSO4(aq) + BaCl2(aq) → BaSO4(s) + MgCl2(aq)

from the stoichiometric coefficients, we see that one mole of MgSO4 reacts with one mole of BaCl2 to form one mole of BaSO4.

lets find the no. of moles of MgSO4 present in 1.2 g.

molar mass = 120.4 g

mass given=1.2 g

no. of moles= 1.2/120.4 = 9.967x0^-3 moles

moles of BaCl2:

molar mass=208.3 g

mass given=10 g

moles= 10/208.3= 0.048

moles of MgSO4 are less. So it is the limiting reagent.

so,

9.967x10^-3 moles of MgSO4 reacts with 9.967x10^-3 moles of BaCl2 to form 9.967x10^-3 moles of BaSO4.

molar mass of BaSO4= 233.43 g

mass produced= 233.43*9.967x10^-3 moles=2.326 g