Answered

Problema 5: ¿Qué volumen de cloro (en CNPT) puede obtenerse si se trata 90 gramos de
MnO2 con suficiente HCI concentrado si en el proceso hay una pérdida del 10%?
MnO2(s) + 4 HCl (ac) → MnCl2(s) + Cl2(g) + 2 H2O (1)



Answer :

GinnyAnswer
Brainly AI Helper here! To determine the volume of chlorine gas (at STP) that can be obtained when 90 grams of MnO2 reacts with excess concentrated HCl with a 10% loss, you can follow these steps: 1. Calculate the molar mass of MnO2: - Mn: 54.94 g/mol - O: 16.00 g/mol (x 2 atoms) - Total molar mass = 54.94 + 2(16.00) = 86.94 g/mol 2. Determine the number of moles of MnO2: - Moles = Mass / Molar mass - Moles = 90 g / 86.94 g/mol ≈ 1.03 mol 3. Use the balanced chemical equation to find the moles of Cl2 produced: - From the balanced equation, 1 mole of MnO2 produces 1 mole of Cl2 - Moles of Cl2 = 1.03 mol 4. Adjust for the 10% loss: - Actual moles of Cl2 produced = Moles of Cl2 x (1 - Loss percentage) - Actual moles of Cl2 = 1.03 mol x (1 - 0.10) = 0.927 mol 5. Calculate the volume of Cl2 gas at STP: - 1 mol of any gas occupies 22.4 L at STP - Volume of Cl2 gas = Moles of Cl2 x 22.4 L/mol - Volume of Cl2 gas = 0.927 mol x 22.4 L/mol ≈ 20.77 L Therefore, the volume of chlorine gas (at STP) that can be obtained is approximately 20.77 liters when 90 grams of MnO2 reacts with excess concentrated HCl with a 10% loss.