Question

Magnesium sulfate reacts with barium chloride according to the following balanced equation. of 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

Answer 1

First write the formula and make sure it is balance

MgSO4 + BaCl2 --> MgCl2 + BaSO4

then, we need to determine which reactant is the limiting reagent. using their respective molar masses.

1.20 g MgSO4 x (1 mole MgSO4/ 120.3 g)=0.00998 moles

10.0 g BaCl2 x (1 mole BaCl/ 172 g) = 0.0580 moles

Since MgSO4 give the less amount of moles, MgSO4 is the limiting reagent. 

0.00998 moles MgSO4 (1 mole BaSO4/1 mole MgSO4) (233.3 g/ 1 mole BaSO4) = 2.34 grams