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%

so whats the answer tell me

option c: 1605.1 kj

the balanced chemical reaction for combustion of methane is as follows:

here, enthalpy formation of is -74.7 kj/mol, 0 kj/mol, -393.5 kj/mol and -241.82 kj/mol respectively.

the heat released or can be calculated as follow:

putting the values,

thus, heat released by combustion of 1 mol of methane is 802.54 kj/mol. thus, for 2 mol, heat released will be:

therefore, heat released by the combustion of 2 mol of methane is 1605.1 kj.