Question

Определить объём водорода, который выделится при взаимодействии 5 г цинка с соляной кислотой.

Answer 1

Ответ

5*22,4/65=1,72

Объяснение:

Answer 2

Calculation of Hydrogen Gas Volume

To determine the volume of hydrogen gas that will be released when 5 grams of zinc reacts with hydrochloric acid, we can use the balanced chemical equation for the reaction:

Zn + 2HCl → H2 + ZnCl2

From the equation, we can see that 1 mole of zinc (Zn) reacts with 2 moles of hydrochloric acid (HCl) to produce 1 mole of hydrogen gas (H2) and 1 mole of zinc chloride (ZnCl2).

To calculate the number of moles of zinc (n(Zn)) in 5 grams of zinc, we can use the formula:

n(Zn) = m(Zn) / M(Zn)

Where: - n(Zn) is the number of moles of zinc - m(Zn) is the mass of zinc (5 grams in this case) - M(Zn) is the molar mass of zinc (65 g/mol)

Using the given values, we can calculate the number of moles of zinc:

n(Zn) = 5 g / 65 g/mol = 0.077 moles of zinc

Since the stoichiometry of the reaction tells us that the number of moles of hydrogen gas produced is equal to the number of moles of zinc, we have:

n(H2) = n(Zn) = 0.077 moles of hydrogen gas

Now, to calculate the volume of hydrogen gas (V(H2)) at standard temperature and pressure (STP), we can use the ideal gas law:

V(H2) = n(H2) * Vm

Where: - V(H2) is the volume of hydrogen gas - n(H2) is the number of moles of hydrogen gas (0.077 moles in this case) - Vm is the molar volume of a gas at STP (22.4 L/mol)

Using the given values, we can calculate the volume of hydrogen gas:

V(H2) = 0.077 moles * 22.4 L/mol = 1.73 L

Therefore, the volume of hydrogen gas that will be released when 5 grams of zinc reacts with hydrochloric acid is approximately 1.73 liters.