Question

In aqueous solution, metal oxides can react with acids to form a salt and water:

Fe₃O₃(s) + 6 HCI(aq)-→ 2 FeCl₃(aq) + 3 H₂O(l)
How many moles of each product will be formed when 35 g of Fe₂O react with 35 g of HCl?

Answer 1

When 35 g of Fe3O3 reacts with 35 g of HCl, approximately 0.438 moles of FeCl3 and 0.657 moles of H2O will be formed.

Step-by-step explanation:

The given equation represents the reaction between Fe3O3(s) and HCl(aq) to form FeCl3(aq) and H2O(l). The stoichiometry of the balanced equation tells us that 1 mole of Fe3O3 reacts with 6 moles of HCl to form 2 moles of FeCl3 and 3 moles of H2O.

To determine the moles of each product formed when 35 g of Fe3O3 reacts with 35 g of HCl, we need to calculate the number of moles of each reactant using their respective molar masses and then use the stoichiometry of the reaction.

Let's calculate the moles of Fe3O3 and HCl:

1. Mass of Fe3O3: 35 g
2. Molar mass of Fe3O3: 159.69 g/mol
3. Moles of Fe3O3 = Mass / Molar mass = 35 g / 159.69 g/mol ≈ 0.219 mol
4. Mass of HCl: 35 g
5. Molar mass of HCl: 36.46 g/mol
6. Moles of HCl = Mass / Molar mass = 35 g / 36.46 g/mol ≈ 0.959 mol

According to the stoichiometry of the reaction, the ratio of reactants to products is 1:6. Thus, for every 1 mole of Fe3O3, 6 moles of HCl are required. However, we have excess HCl (0.959 mol) and limiting Fe3O3 (0.219 mol).

Since Fe3O3 is the limiting reactant, the moles of FeCl3 and H2O formed will be determined by the stoichiometry of Fe3O3 and the ratio of products to Fe3O3, which is 2:1 for FeCl3 and 3:1 for H2O.

Moles of FeCl3 formed = 2 × Moles of Fe3O3 = 2 × 0.219 mol = 0.438 mol

Moles of H2O formed = 3 × Moles of Fe3O3 = 3 × 0.219 mol = 0.657 mol