Aluminum and chlorine combine to form aluminum chloride. (7.6,7.7,7.8) 2 Al(s)+3

a. From the balanced chemical equation, we can see that 2 moles of Al react with 3 moles of Cl$_

Question

Aluminum and chlorine combine to form aluminum chloride. $$\begin{aligned}&(7.6,7.7,7.8) \\ &\quad 2 \mathrm{Al}(s)+3 \mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{AlCl}_{3}(s) \end{aligned}$$ a. How many moles of $\mathrm{Cl}_{2}$ are needed to react with $2.50$ moles of $\mathrm{Al}$ ? b. How many grams of $\mathrm{AlCl}_{3}$ are produced when $37.7 \mathrm{~g}$ of Al reacts? c. When $13.5 \mathrm{~g}$ of Al reacts with $8.00 \mathrm{~g}$ of $\mathrm{Cl}_{2}$, how many grams of $\mathrm{AlCl}_{3}$ can form? d. If $45.0 \mathrm{~g}$ of $\mathrm{Al}$ and $62.0 \mathrm{~g}$ of $\mathrm{Cl}_{2}$ are mixed, and $66.5 \mathrm{~g}$ of $\mathrm{AlCl}_{3}$ is actually obtained, what is the percent yield of $\mathrm{AlCl}_{3}$ for the reaction?

          Aluminum and chlorine combine to form aluminum chloride.
$$\begin{aligned}&(7.6,7.7,7.8) \\
&\quad 2 \mathrm{Al}(s)+3 \mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{AlCl}_{3}(s)
\end{aligned}$$
a. How many moles of $\mathrm{Cl}_{2}$ are needed to react with $2.50$ moles of $\mathrm{Al}$ ?
b. How many grams of $\mathrm{AlCl}_{3}$ are produced when $37.7 \mathrm{~g}$ of Al reacts?
c. When $13.5 \mathrm{~g}$ of Al reacts with $8.00 \mathrm{~g}$ of $\mathrm{Cl}_{2}$, how many grams of $\mathrm{AlCl}_{3}$ can form?
d. If $45.0 \mathrm{~g}$ of $\mathrm{Al}$ and $62.0 \mathrm{~g}$ of $\mathrm{Cl}_{2}$ are mixed, and $66.5 \mathrm{~g}$ of $\mathrm{AlCl}_{3}$ is actually obtained, what is the percent yield of $\mathrm{AlCl}_{3}$ for the reaction?

Added by Crystal P.

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