How much CuSO4 will be formed if 200 g of copper hydroxide with 300 g of 75% sulfuric acid?

To answer this question, we need to use stoichiometry, which is a section of chemistry that involves using relationships between reactants and/or products in a chemical reaction to determine desired quantitative data.

The balanced chemical equation for the reaction between copper hydroxide and sulfuric acid is:

Cu(OH)2 + H2SO4 → CuSO4 + 2H2O

From this equation, we can see that one mole of copper hydroxide reacts with one mole of sulfuric acid to produce one mole of copper sulfate.

Step 1: Calculate the number of moles of copper hydroxide and sulfuric acid.

The molar mass of copper hydroxide (Cu(OH)2) is approximately 97.56 g/mol. So, the number of moles of copper hydroxide is 200 g / 97.56 g/mol = 2.05 mol.

The molar mass of sulfuric acid (H2SO4) is approximately 98.08 g/mol. However, we only have 75% of this, so the actual mass of sulfuric acid we have is 300 g * 0.75 = 225 g. So, the number of moles of sulfuric acid is 225 g / 98.08 g/mol = 2.29 mol.

Step 2: Determine the limiting reactant.

From the balanced chemical equation, we know that one mole of copper hydroxide reacts with one mole of sulfuric acid. Since we have more moles of sulfuric acid than copper hydroxide, copper hydroxide is the limiting reactant.

Step 3: Calculate the mass of copper sulfate produced.

From the balanced chemical equation, we know that one mole of copper hydroxide produces one mole of copper sulfate. So, we will produce 2.05 mol of copper sulfate.

The molar mass of copper sulfate (CuSO4) is approximately 159.61 g/mol. So, the mass of copper sulfate produced is 2.05 mol * 159.61 g/mol = 327.2 g.

So, if 200 g of copper hydroxide reacts with 300 g of 75% sulfuric acid, 327.2 g of copper sulfate will be formed.