if aluminum chloride and copper sulfate make 48g of aluminum sulfate how many grams of copper sulfate was used ?
To find the number of grams of copper sulfate used, we need to calculate the mass of copper sulfate that reacts with aluminum chloride to form 48 grams of aluminum sulfate.
First, we can write down the balanced chemical equation for the reaction between aluminum chloride and copper sulfate:
2AlCl₃ + 3CuSO₄ → Al₂(SO₄)₃ + 3CuCl₂
From the balanced equation, we can see that the molar ratio between copper sulfate (CuSO₄) and aluminum sulfate (Al₂(SO₄)₃) is 3:1. This means that for every 3 moles of copper sulfate, we get 1 mole of aluminum sulfate.
To calculate the number of moles of aluminum sulfate, we need to use its molar mass. The molar mass of aluminum sulfate (Al₂(SO₄)₃) can be calculated by adding up the molar masses of each element:
2(Al) + 3(S) + 12(O) = 342 g/mol
Next, we need to calculate the number of moles of aluminum sulfate. We can use the formula:
Number of moles = Mass / Molar mass
Given that the mass of aluminum sulfate is 48 grams and the molar mass is 342 g/mol, we have:
Number of moles of aluminum sulfate = 48 g / 342 g/mol ≈ 0.1408 mol
Since the molar ratio between copper sulfate and aluminum sulfate is 3:1, we can now calculate the number of moles of copper sulfate used:
Number of moles of copper sulfate = 0.1408 mol x (3 mol CuSO₄ / 1 mol Al₂(SO₄)₃)
≈ 0.4224 mol
Finally, to find the mass of copper sulfate used, we can multiply the number of moles by its molar mass. The molar mass of copper sulfate (CuSO₄) is:
63.5 (Cu) + 32.1 (S) + 4(16.0) (O) = 159.5 g/mol
Mass of copper sulfate used = Number of moles x Molar mass
= 0.4224 mol x 159.5 g/mol
≈ 67.3 grams
Therefore, approximately 67.3 grams of copper sulfate was used in the reaction to produce 48 grams of aluminum sulfate.