Question

який об'єм водню (н.у) утвориться під час взаємодії 130г цинку з хлоридною кислотою? ПОМОГИТЕ ПЖЖЖЖЖ СРОЧНО НУЖНО​

Answer 1

Zn + 2HCl → ZnCl2 + H2

m / M = V / Vm

130 / 65 = V / 22,4

V = 130 × 22,4 / 65 = 44,8 л

Answer 2

Calculation of Hydrogen Gas Volume

To calculate the volume of hydrogen gas produced when 130g of zinc reacts with hydrochloric acid, we can use the balanced chemical equation for the reaction between zinc and hydrochloric acid:

Zn + 2HCl → ZnCl2 + H2

According to the stoichiometry of the balanced chemical equation, 1 mole of zinc 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 moles of zinc: - The molar mass of zinc (Zn) is approximately 65.38 g/mol. - Therefore, the number of moles of zinc can be calculated using the formula: moles = mass / molar mass moles = 130g / 65.38 g/mol moles ≈ 1.99 moles

2. Calculate the moles of hydrogen gas: - Since the stoichiometry of the balanced chemical equation is 1:1 for zinc and hydrogen gas, the moles of hydrogen gas produced will be the same as the moles of zinc used. - Therefore, the moles of hydrogen gas produced ≈ 1.99 moles.

3. Calculate the volume of hydrogen gas: - The volume of 1 mole of any gas at standard temperature and pressure (STP) is approximately 22.4 liters. - Therefore, the volume of hydrogen gas produced can be calculated using the formula: volume = moles * 22.4 L/mol volume ≈ 1.99 moles * 22.4 L/mol volume ≈ 44.576 L

Conclusion

When 130g of zinc reacts with hydrochloric acid, approximately 44.576 liters of hydrogen gas will be produced.