Question

How many grams of aluminum (Al) would react completely with 1350 grams of copper (II) chloride (CuCl2) according to the following equation?

2Al + 3CuCl2 ---> 2AlCl3 + 3Cu

Answer 1

Answer on Question #50325, Chemistry, Other

**Task:**

How many grams of aluminum (Al) would react completely with 1350 grams of copper (II) chloride (CuCl₂) according to the following equation?

$$2\text{Al} + 3\text{CuCl}_2 \rightarrow 2\text{AlCl}_3 + 3\text{Cu}$$

**Answer:**

$$\nu = \frac{m}{M}$$ $$M(Al) = 27\,g/mol$$ $$M(\text{CuCl}_2) = 134.5\,g/mol$$ $$\nu(\text{CuCl}_2) = \frac{1350}{134.5} = 10.04\,mol$$ $$\nu(\text{Al}) = \frac{2 \cdot \nu(\text{CuCl}_2)}{3} = 6.7\,mol$$ $$m(\text{Al}) = \nu(\text{Al}) \cdot M(\text{Al})$$ $$m(\text{Al}) = 6.7 \cdot 27 = 180.7\,g$$

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