Question

При взаимодействии цинка с раствором ортофосфорной кислоты выделилось 0,224 л водорода (н. у.). Определите массу цинка, прореагировавшего с кислотой.

Answer 1

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Answer 2

Interaction of Zinc with Orthophosphoric Acid

When zinc reacts with orthophosphoric acid (H₃PO₄), hydrogen gas (H₂) is produced. The volume of hydrogen gas released during the reaction is given as 0.224 L at standard temperature and pressure (n.t.p).

To determine the mass of zinc that reacted with the acid, we can use the ideal gas law and stoichiometry. The balanced chemical equation for the reaction between zinc and orthophosphoric acid is:

Zn + 2H₃PO₄ → Zn(H₂PO₄)₂ + H₂

From the balanced equation, we can see that 1 mole of zinc reacts with 2 moles of orthophosphoric acid to produce 1 mole of hydrogen gas.

To calculate the mass of zinc, we need to convert the volume of hydrogen gas to moles using the ideal gas law. At n.t.p, 1 mole of any ideal gas occupies 22.4 L. Therefore, the number of moles of hydrogen gas can be calculated as:

moles of H₂ = volume of H₂ (in L) / 22.4 L/mol

Substituting the given volume of hydrogen gas (0.224 L) into the equation, we get:

moles of H₂ = 0.224 L / 22.4 L/mol = 0.01 mol

Since the stoichiometry of the reaction tells us that 1 mole of zinc reacts with 1 mole of hydrogen gas, the number of moles of zinc is also 0.01 mol.

To determine the mass of zinc, we need to use its molar mass. The molar mass of zinc (Zn) is approximately 65.38 g/mol Therefore, the mass of zinc that reacted with the acid can be calculated as:

mass of zinc = moles of Zn × molar mass of Zn

mass of zinc = 0.01 mol × 65.38 g/mol = 0.6538 g

Therefore, the mass of zinc that reacted with the orthophosphoric acid is approximately 0.6538 grams.