Question

Рассчитайте объем водорода образовавшегося при взаимодействии 54г алюминия с соляной кислотой

Answer 1

Если что-то не понятно спрашивай

Answer 2

Calculating the Volume of Hydrogen Gas Produced

To calculate the volume of hydrogen gas produced when 54g of aluminum reacts with hydrochloric acid, we can use the following chemical equation:

2 Al + 6 HCl → 2 AlCl3 + 3 H2

According to the stoichiometry of the balanced chemical equation, 2 moles of aluminum react with 6 moles of hydrochloric acid to produce 3 moles of hydrogen gas.

Calculation Steps

1. Calculate the moles of aluminum using its molar mass. 2. Use the mole ratio from the balanced chemical equation to find the moles of hydrogen gas produced. 3. Convert the moles of hydrogen gas to volume using the ideal gas law.

Calculation

1. Moles of Aluminum: - The molar mass of aluminum (Al) is 27 g/mol. - Moles of aluminum = Mass of aluminum / Molar mass of aluminum - Moles of aluminum = 54g / 27 g/mol - Moles of aluminum = 2 moles

2. Moles of Hydrogen Gas: - Using the mole ratio from the balanced chemical equation: - 2 moles of aluminum produce 3 moles of hydrogen gas. - Moles of hydrogen gas = (2 moles of aluminum) * (3 moles of H2 / 2 moles of Al) - Moles of hydrogen gas = 3 moles

3. Volume of Hydrogen Gas: - Using the ideal gas law: PV = nRT, where P is pressure, V is volume, n is moles, R is the gas constant, and T is temperature. - Assuming standard temperature and pressure (STP) conditions (0°C and 1 atm pressure), the molar volume of a gas is 22.4 L/mol. - Volume of hydrogen gas = Moles of hydrogen gas * Molar volume at STP - Volume of hydrogen gas = 3 moles * 22.4 L/mol - Volume of hydrogen gas = 67.2 L

Conclusion

When 54g of aluminum reacts with hydrochloric acid, approximately 67.2 liters of hydrogen gas is produced at standard temperature and pressure (STP).