Question

5.0 g of Ag2CO3(aq) is reacted with excess KBr(aq) to produce 4.3 g of AgBr(s). What is the percent yield of AgBr in the reaction

Answer 1

The percentage yield of AgBr is 63.14%

We'll begin by writing the balanced equation for the reaction. This is given below:

Ag₂CO₃ + 2KBr —> 2AgBr + K₂CO₃

Molar mass of Ag₂CO₃ = (2×108 ) + 12 + (3×16) = 276 g/mol

Mass of Ag₂CO₃ from the balanced equation = 1 × 276 = 276 g

Molar mass of AgBr = 108 + 80 = 188 g/mol

Mass of AgBr from the balanced equation = 2 × 188 = 376 g

From the balanced equation above,

276 g of Ag₂CO₃ reacted to produce 376 g of AgBr

• Next, we shall determine the theoretical yield of AgBr

From the balanced equation above,

276 g of Ag₂CO₃ reacted to produce 376 g of AgBr

Therefore,

5 g of Ag₂CO₃ will react to produce = (5 × 376) / 276 = 6.81 g of AgBr

Thus, the theoretical yield of AgBr is 6.81 g

• Finally, we shall determine the percentage yield of AgBr.

Actual yield of AgBr = 4.3 g

Theoretical yield of AgBr = 6.81 g

Percentage yield =?

$percentage \: yield \: = \frac{actual}{theoretical} \times 100 \\ \\ percentage \: yield \: = \frac{4.3}{6.81} \times 100 \\ \\ percentage \: yield \: = 63.14\%$

Thus, the percentage yield of AgBr is 63.14%

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