Question

К 100граммам 9,8℅ раствора H2SO4 добавили магнии.Определите V(H2)выделившегося при этом!

Answer 1

Сколько магния добавили?

Answer 2

Calculating the Volume of Hydrogen Gas Produced

To calculate the volume of hydrogen gas produced when magnesium is added to a 100g 9.8% H2SO4 solution, we can use the chemical equation for the reaction between magnesium and sulfuric acid:

Mg + H2SO4 → MgSO4 + H2

The balanced chemical equation shows that 1 mole of magnesium reacts with 1 mole of sulfuric acid to produce 1 mole of hydrogen gas.

Given that the molar mass of magnesium is approximately 24.305 g/mol and the molar mass of sulfuric acid is approximately 98.079 g/mol, we can proceed with the calculation.

First, we need to find the number of moles of sulfuric acid in the solution:

Number of moles of H2SO4 = (mass of H2SO4) / (molar mass of H2SO4)

Number of moles of H2SO4 = (9.8g * 0.01) / 98.079g/mol

Number of moles of H2SO4 ≈ 0.01 moles

Since the reaction between magnesium and sulfuric acid occurs in a 1:1 mole ratio, the number of moles of hydrogen gas produced will be the same as the number of moles of sulfuric acid.

Therefore, the volume of hydrogen gas produced can be calculated 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)

Given that the reaction likely occurs at standard temperature and pressure (STP), which is 0°C (273.15 K) and 1 atm, we can simplify the calculation.

V = nRT / P

Substituting the known values: V = (0.01 moles) * (0.0821 L.atm/mol.K) * (273.15 K) / 1 atm

V ≈ 0.224 L

So, the volume of hydrogen gas produced when magnesium is added to the 100g 9.8% H2SO4 solution is approximately 0.224 liters.