Question

Assuming 34.51 grams of Al are consumed in the presence of excess copper II chloride dihydrate, how many grams of AlCl3 can be produced if the reaction will only produce 86.52% yield?

Answer 1

Answer:

147 g of $AlCl_{3}$ are produced with 86.52% yield.

Explanation:

1. Write the balanced chemical reaction for the aluminum consumed in the presence of copper II chloride dihydrate:

$2Al+3Cl_{2}CuH_{4}O_{2}=2Cu+2AlCl_{3}+6H_{2}O$

2. Calculate the maximum quantity of $AlCl_{3}$ that can be produced:

The limiting reagent is the Al, because the problem says that there are excess of copper II chloride dihydrate $Cl_{2}CuH_{4}O_{2}$

$34.51gAl*\frac{1molAl}{27gAl}*\frac{2molesAlCl_{3}}{2molesAl}*\frac{133gAlCl_{3}}{1molAlCl_{3}}=170gAlCl_{3}$

3. Calculate the quantity of $AlCl_{3}$ produced ith 86.52% yield:

$170gAlCl_{3}*\frac{86.52}{100}=147gAlCl_{3}$