A solution containing 4.15 g of magnesium chloride is added to a second solution containing excess sodium sulfate. Magnesium sulfate precipitates.
•Calculate the ideal yield of magnesium sulfate
. If actual yield is 3.95 g, calculate percentage yield
1. Eq.:
2. Given:
3. Wanted:
4.

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%

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question: ray weighs 100 kilograms, his wrestling partner, bobby, weighs 50 kilograms. if ray moves at 4 m s , how fast will bobby have to run at him to apply an equal amount of kinetic energy?

answer: 8 m/s he needs to run twice as fast to keep up so he has some work to do

explanation:   if the mass has units of kilograms and the velocity of meters per second, the kinetic energy has units of kilograms-meters squared per second squared.

question answered by

(jacemorris04)