Question

Какой объем водорода выделится при взаимодействии Mg с 400 г 14,6%-ой соляной кислоты?

Answer 1

Mg+2HCl=MgCl2+H2 m в-ва(HCl)=m р-ра* w=146*0,1=14,6 г n(HCL)=m/M=14,6/36,5=0,4 моль n(HCL):n(H2) 2:1 0,4:x-------->x=0,2 моль V(H2)=n*Vm=0,2*22,4= 4,48 л Ответ: объем Н2 равен 4,48 л

Answer 2

Reaction and Calculation

To calculate the volume of hydrogen gas produced when magnesium reacts with 400g of 14.6% hydrochloric acid, we can use the balanced chemical equation for the reaction between magnesium and hydrochloric acid:

Mg + 2HCl → MgCl2 + H2

According to the stoichiometry of the balanced chemical equation, 1 mole of magnesium reacts with 2 moles of hydrochloric acid to produce 1 mole of hydrogen gas. We can use this information to calculate the volume of hydrogen gas produced.

Calculation Steps

1. Calculate the number of moles of hydrochloric acid. 2. Use the mole ratio from the balanced chemical equation to find the number of moles of hydrogen gas produced. 3. Convert the moles of hydrogen gas to volume using the ideal gas law.

Calculation

1. Calculate the number of moles of hydrochloric acid: - The molar mass of HCl is approximately 36.46 g/mol. - The percentage concentration of the hydrochloric acid is 14.6%. - Therefore, the mass of HCl in 400g of 14.6% HCl solution is 400g * 0.146 = 58.4g. - The number of moles of HCl is 58.4g / 36.46g/mol ≈ 1.60 moles.

2. Use the mole ratio to find the number of moles of hydrogen gas produced: - According to the balanced chemical equation, 1 mole of magnesium produces 1 mole of hydrogen gas. - Therefore, the number of moles of hydrogen gas produced is also approximately 1.60 moles.

3. Convert the moles of hydrogen gas to volume using the ideal gas law: - The ideal gas law is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. - Assuming standard temperature and pressure (STP), where T = 273.15 K and P = 1 atm, the volume of 1 mole of any gas is 22.4 liters. - Therefore, the volume of hydrogen gas produced is approximately 1.60 moles * 22.4 liters/mole = 35.84 liters.

Conclusion

The volume of hydrogen gas produced when magnesium reacts with 400g of 14.6% hydrochloric acid is approximately 35.84 liters.