Question

use the following information to answer the next question. according to the equation: 3cacl₂ + al₂(so₄)₃→2alcl₃ + 3caso₄ 9. how many moles of alcl₃ will be formed when 3.54 moles of cacl₂ react? 2.05 moles 5.31 moles 2.36 moles 4.60 moles

Explanation:

Step1: Identify mole - ratio

From the balanced equation $3CaCl_2 + Al_2(SO_4)_3
ightarrow2AlCl_3+3CaSO_4$, the mole - ratio of $CaCl_2$ to $AlCl_3$ is 3:2.

Step2: Set up proportion

Let $x$ be the number of moles of $AlCl_3$ formed. We have the proportion $\frac{3\ mol\ CaCl_2}{2\ mol\ AlCl_3}=\frac{3.54\ mol\ CaCl_2}{x}$.

Step3: Solve for $x$

Cross - multiply: $3x = 2\times3.54$. Then $x=\frac{2\times3.54}{3}=\frac{7.08}{3}=2.36$ moles.

Answer:

2.36 moles