Question

Mr. Ragusa asks Hassan to make silver crystals from the following reaction.

2AgNO3 + Cu --> Cu(NO3)2 + 2Ag
Mr. Ragusa gives Hassan 51 g of silver nitrate to conduct the experiment. Hassan gives Mr. Ragusa 20 g of
silver crystals. What was Hassan's percent yield for his experiment?
Please round your answer to 2 decimal places

Answer 1

Percentage yield = 61.7%

Step-by-step explanation:

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

2AgNO₃ + Cu —> Cu(NO₃)₂ + 2Ag

Next, we shall determine the mass of AgNO₃ that reacted and the mass of Ag produced from the balanced equation. This is illustrated below:

Molar mass of AgNO₃ = 108 + 14 + (16×3)

= 108 + 14 + 48

= 170 g/mol

Mass of AgNO₃ from the balanced equation = 2 × 170 = 340 g

Molar mass of Ag = 108 g/mol

Mass of Ag from the balanced equation = 2 × 108 = 216 g

SUMMARY:

From the balanced equation above,

340 g of AgNO₃ reacted to produce 216 g of Ag.

Next, we shall determine the theoretical yield of Ag. This can be obtained as follow:

From the balanced equation above,

340 g of AgNO₃ reacted to produce 216 g of Ag.

Therefore, 51 g of AgNO₃ will react to produce = (51 × 216)/340 = 32.4 g of Ag.

Thus, the theoretical yield of Ag is 32.4 g.

Finally, we shall determine the percentage yield of Ag. This can be obtained as follow:

Actual yield = 20 g

Theoretical yield = 32.4 g

Percentage yield =?

Percentage yield = Actual yield /Theoretical yield × 100

Percentage yield = 20 / 32.4 × 100

Percentage yield = 61.7%