Hi, I would like some help with my chemistry problem.

Question

The question is: Consider the double replacement reaction between magnesium sulfate and potassium bromide. If a reaction vessel contains 5.02g of magnesium sulfate and 9.89g of potassium bromide, how many grams of each product are produced?

Answer 1

Magnesium sulfate = MgSO₄ (molar mass = 120.3 g/mol)

Potassium bromide = KBr (molar mass = 119.0 g/mol)

MgSO₄(aq) + 2KBr(aq) ==> MgBr₂(aq) + K₂SO₄(aq)

This would normally be a double replacement but since ALL species are aqueous (aq), THERE IS NO REACTION. But, since the question is asking you to write an answer to the "theoretical reaction", I will provide such an answer. (Please mention to your teacher that there is no reaction taking place)

5.02 g MgSO₄ x 1 mol / 120.3 g = 0.0417 mols MgSO₄

9.89 g KBr x 1 mol / 119 g = 0.0831 mols KBr

Since they are in a 1:1 mole ratio in the balanced equation, they are both present in stoichiometric equivalent amounts (2 x 0.0417 = 0.0834 ~= 0.0831). Thus, neither reactant is limiting.

Grams MgBr₂ formed = 0.0831 mol KBr x 1 mol MgBr₂ / 2 mol KBr x 184 g/mol = 7.65 g MgBr₂

Grams K₂SO₄ formed = 0.0831 mol KBr x 1 mol K₂SO₄ / mol KBr x 174 g/ mol = 14.5 g K₂SO₄