Question

Какой объем водорода (н.у) выделится при взаимодействии 13 г цинка с соляной кислотой?

Answer 1

Zn+2Hcl=ZnCl2+H2
13 гр
n(Zn) =13/65=0.2 моль
n(Zn) =2n (H2) по уравнению
n(H2) =0.4 моль
V (H2)=0.4*22.4=8.96 л

Answer 2

Hydrogen Gas Volume Calculation

To calculate the volume of hydrogen gas produced when 13g of zinc reacts with hydrochloric acid, we can use the chemical equation for the reaction:

Zn + 2HCl → H2 + ZnCl2

According to the stoichiometry of the reaction, 1 mole of zinc reacts with 2 moles of hydrochloric acid to produce 1 mole of hydrogen gas. The molar mass of zinc (Zn) is 65 g/mol.

Using this information, we can calculate the volume of hydrogen gas produced.

The calculation involves the following steps:

1. Calculate the number of moles of zinc (n(Zn)) involved in the reaction. 2. Use the stoichiometry of the reaction to find the number of moles of hydrogen gas (n(H2)) produced. 3. Finally, calculate the volume of hydrogen gas produced using the ideal gas law.

Calculation Steps

1. Calculate the number of moles of zinc (n(Zn)): - n(Zn) = m(Zn) / M(Zn) = 13g / 65 g/mol = 0.2 moles.

2. Use stoichiometry to find the number of moles of hydrogen gas (n(H2)) produced: - According to the balanced chemical equation, the molar ratio of Zn to H2 is 1:1. Therefore, n(H2) = n(Zn) = 0.2 moles.

3. Calculate the volume of hydrogen gas produced using the ideal gas law: - The ideal gas law is given by the equation: 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. - At standard temperature and pressure (STP), the volume of 1 mole of any gas is 22.4 liters. - Therefore, the volume of hydrogen gas produced is V(H2) = n(H2) * 22.4 L/mol = 0.2 moles * 22.4 L/mol = 4.48 liters.

Therefore, the volume of hydrogen gas (at standard temperature and pressure) produced when 13g of zinc reacts with hydrochloric acid is 4.48 liters.