Question

Silver nitrate reacts with aluminum chloride to form the insoluble compound, silver chloride. The reaction proceeds according to the balanced equation below: 3 AgNO3 (aq) + 1 AICIz (aq) + 1 Al(NO3), (aq) + 3 AgCl (s) Wayne reacted 1.616 g of aluminum chloride with an excess amount of silver nitrate. Based on the balanced chemical equation above and the mass of aluminum chloride, determine Wayne's theoretical yield (in grams) of solid silver chloride. Molar mass silver nitrate: 169.87 g/mol Molar mass aluminum chloride: 133.34 g/mol Molar mass silver chloride: 143.32 g/mol Note: Do not use scientific notation or units in your response. Sig figs will not be graded in this question, enter your response to four decimal places. Carmen may add or remove digits from your response, your submission will still be graded correctly if this happens.

Answer 1

The theoretical yield of silver chloride, AgCl is 5.2109 g

The balanced equation for the reaction is given below:

3AgNO₃(aq) + AICI₃(aq) —> Al(NO₃)₃ (aq) + 3AgCl (s)

Next, we shall determine the mass of aluminum chloride, AICI₃ that reacted and the mass of silver chloride, AgCl produced from the balanced equation. This is illustrated below:

Molar mass of AICI₃ = 133.34 g/mol

Mass of AICI₃ from the balanced equation = 1 × 133.34 = 133.34 g

Molar mass of AgCl = 143.32 g/mol

Mass of AgCl from the balanced equation = 3 × 143.32 = 429.96 g

SUMMARY

From the balanced equation above,

133.34 g of AICI₃ reacted to produce 429.96 g of AgCl.

Finally, we shall determine the theoretical yield of AgCl by the reaction of 1.616 g of AICI₃ as follow:

From the balanced equation above,

133.34 g of AICI₃ reacted to produce 429.96 g of AgCl.

Therefore,

1.616 g of AICI₃ will react to produce = $\frac{1.616 * 429.96}{133.34}$ = 5.2109 g of AgCl.

Thus, the theoretical yield of silver chloride, AgCl is 5.2109 g

Learn more: brainly.com/question/24653699