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uzmaali3 uzmaali3
wrote...
Posts: 1
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7 years ago
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?
 2.08 g
 1.20 g
 10.0 g
 2.33 g
 23.3 g
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Replies
wrote...
Subject Expert
7 years ago
Hi,

MgSO4 + BaCl2 → MgCl2 + BaSO4

Assuming MgSO4 is excess: (10/208.23)*233.43 =  11.2 g BaSO4
Assuming BaCl2 is excess : (1.2/120.366)*233.43 = 2.33 g BaSO4

You can get maximum of 2.33g BaSO4 with the stated conditions.

Hope this helps,
wrote...
Valued Member
7 years ago
MgSO4(aq) + BaCl2(aq) → MgCl2(aq) + BaSO4(s)

Complete ionic equation:
Mg^2+(aq) + SO4^2-(aq) + Ba^2+(aq) + 2 Cl^-(aq) → Mg^2+(aq) + 2 Cl^-(aq) + BaSO4(s)

Net ionic equation:
Ba^2+(aq) + SO4^2-(aq) → BaSO4(s)

Answer=2.33g BaSO4
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