2al + 3cl₂ → 2alcl₃\nhow many grams of aluminum will react fully with 1.25 moles cl₂?\nmolar mass al: 26.98…

Answer:

20.0 g

Explanation:

Step1: Determine mole - ratio

From the balanced equation $2Al + 3Cl_2\rightarrow2AlCl_3$, the mole - ratio of $Al$ to $Cl_2$ is $\frac{2}{3}$.

Step2: Calculate moles of Al

Given $n_{Cl_2}=1.25$ mol. Using the mole - ratio, $n_{Al}=\frac{2}{3}n_{Cl_2}=\frac{2}{3}\times1.25$ mol$=\frac{2.5}{3}$ mol.

Step3: Calculate mass of Al

Using the formula $m = nM$, where $M_{Al}=26.98$ g/mol and $n_{Al}=\frac{2.5}{3}$ mol. So $m_{Al}=\frac{2.5}{3}$ mol$\times26.98$ g/mol$\approx20.0$ g.