Question

PLEASE HELP!!!

4.65 mol AlBr3 reacts with 6.25 mol K2SO4 according to the equation below... how many moles aluminum sulfate form from 6.25 mol k2so4

Answer 1

2.08 mol Al₂(SO₄)₃

Step-by-step explanation:

The question provided is a stoichiometry problem, asking for the amount of aluminum sulfate formed from a given amount of potassium sulfate K₂SO₄ when it reacts with aluminum bromide AlBr₃. We will determine the number of moles of aluminum sulfate based on the chemical reaction provided using stoichiometry.

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Given balanced chemical equation:

`2AlBr_3 + 3K_2SO_4 \longrightarrow 6KBr + Al_2(SO_4)_3`

This reaction shows that 2 moles of aluminum bromide (AlBr₃) react with 3 moles of potassium sulfate (K₂SO₄) to produce 6 moles of potassium bromide (KBr) and 1 mole of aluminum sulfate (Al₂(SO₄)₃).

From the balanced chemical equation, we can see that the mole ratio of K₂SO₄ to Al₂(SO₄)₃ is 3:1. Therefore, to find the moles of Al₂(SO₄)₃ produced from 6.25 moles of K₂SO₄, we use the following ratio:

`\Longrightarrow \frac{6.25 \text{ mol $K_2SO_4$}}{}* \frac{1 \text{ mol $Al_2(SO_4)_3$}}{3 \text{ mol $K_2SO_4$}} = \boxed{2.08 \text{ mol $Al_2(SO_4)_3$}}`

Thus, the number of moles of aluminum sulfate formed, is approximately 2.08 moles.