Answer:
1. 5.24 g
2. 75.4%
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
MgCl₂ + Na₂SO₄ —> MgSO₄ + 2NaCl
Next, we shall determine the mass of MgCl₂ that reacted and the mass of MgSO₄ produced from the balanced equation. This can be obtained as follow:
Molar mass of MgCl₂ = 24 + (35.5×2)
= 24 + 71
= 95 g/mol
Mass of MgCl₂ from the balanced equation = 1 × 95 = 95 g
Molar mass of MgSO₄ = 24 + 32 + (16×4)
= 24 +32 + 64
= 120 g/mol
Mass of MgSO₄ from the balanced equation = 1 × 120 = 120 g
SUMMARY:
From the balanced equation above,
95 g of MgCl₂ reacted to produce 120 g of MgSO₄.
1. Determination of the ideal yield of magnesium sulfate, MgSO₄.
From the balanced equation above,
95 g of MgCl₂ reacted to produce 120 g of MgSO₄.
Therefore, 4.15 g of MgCl₂ will react to produce = (415 × 120)/95 = 5.24 g of MgSO₄.
Thus, the ideal yield of magnesium sulfate, MgSO₄ is 5.24 g
2. Determination of the percentage yield.
Actual yield of MgSO₄ = 3.95 g
Ideal yield of MgSO₄ = 5.24 g
Percentage yield =?
Percentage yield = Actual yield / Ideal yield × 100
Percentage yield = 3.95 / 5.24 × 100
Percentage yield = 395 / 5.24
Percentage yield of MgSO₄ = 75.4%
