2al + 6hcl → 3h₂ + 2alcl₃\nhow many grams of alcl₃ form from 7.8 moles of hcl?\nstep 1: how many moles of…

Explanation:

Step1: Determine mole - ratio

From the balanced chemical equation $2Al + 6HCl\rightarrow3H_2 + 2AlCl_3$, the mole - ratio of $HCl$ to $AlCl_3$ is $6:2$ or $3:1$.

Step2: Calculate moles of $AlCl_3$

Let $x$ be the moles of $AlCl_3$. Using the mole - ratio, $\frac{x}{7.8}=\frac{2}{6}$. Solving for $x$, we get $x=\frac{2\times7.8}{6}=2.6$ mol.

Step3: Calculate mass of $AlCl_3$

The molar mass of $AlCl_3$ is $M = 26.98+3\times35.45=26.98 + 106.35=133.33$ g/mol. The mass $m$ of $AlCl_3$ is $m = n\times M$, where $n = 2.6$ mol and $M=133.33$ g/mol. So $m=2.6\times133.33 = 346.66$ g.

Answer:

346.66 g