Question

Какой обьем водорода выделится при взаимодействии 56гр кальция с серной кислотой если выход реакции 95%

Answer 1

Решается таким образом

Answer 2

Reaction and Calculation of Hydrogen Gas Volume

When 56g of calcium reacts with sulfuric acid, the reaction can be represented by the following equation: Ca + H2SO4 → CaSO4 + H2

According to the stoichiometry of the balanced chemical equation, 1 mole of calcium reacts with 1 mole of sulfuric acid to produce 1 mole of hydrogen gas. The molar mass of calcium is 40.08 g/mol, and the molar mass of sulfuric acid (H2SO4) is 98.08 g/mol.

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

First, we need to find the number of moles of calcium: Number of moles = Mass / Molar mass Number of moles = 56g / 40.08 g/mol Number of moles ≈ 1.396 moles

Since the reaction is 95% efficient, we can calculate the volume of hydrogen gas produced using the ideal gas law: PV = nRT

Where: P = pressure (atm) V = volume (L) n = number of moles R = ideal gas constant (0.0821 L.atm/mol.K) T = temperature (K)

Assuming standard temperature and pressure (STP), which is 0°C (273.15 K) and 1 atm, the volume of 1 mole of any gas at STP is 22.4 L.

Therefore, the volume of hydrogen gas produced can be calculated as follows: Volume = Number of moles * 22.4 L Volume ≈ 1.396 moles * 22.4 L Volume ≈ 31.26 L

So, approximately 31.26 liters of hydrogen gas will be produced when 56g of calcium reacts with sulfuric acid, assuming a 95% yield.

Conclusion

When 56g of calcium reacts with sulfuric acid, approximately 31.26 liters of hydrogen gas will be produced, given a 95% yield.