Question

Помогите пожалуйста. Определить m и V газа, полученного при взаимодействии Mg с 49г H2SO4.

Answer 1

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Answer 2

Calculating the Volume and Amount of Gas Produced

To determine the volume and amount of gas produced when magnesium (Mg) reacts with 49g of sulfuric acid (H2SO4), we can use the balanced chemical equation for the reaction between magnesium and sulfuric acid:

Mg + H2SO4 → MgSO4 + H2

From the balanced chemical equation, we can see that 1 mole of magnesium reacts with 1 mole of sulfuric acid to produce 1 mole of hydrogen gas.

Calculating the Amount of Hydrogen Gas Produced (m)

To calculate the amount of hydrogen gas produced (m), we can use the given mass of sulfuric acid (H2SO4) and the molar mass of sulfuric acid to find the number of moles of sulfuric acid. Then, we can use the stoichiometry of the balanced chemical equation to find the number of moles of hydrogen gas produced.

The molar mass of sulfuric acid (H2SO4) is: - Hydrogen (H): 1.007 g/mol - Sulfur (S): 32.06 g/mol - Oxygen (O): 16.00 g/mol So, the molar mass of H2SO4 = 2(1.007) + 32.06 + 4(16.00) = 98.08 g/mol

Using the given mass of sulfuric acid (H2SO4): 49g / 98.08 g/mol = 0.499 moles of H2SO4

Since 1 mole of sulfuric acid reacts to produce 1 mole of hydrogen gas, the amount of hydrogen gas produced (m) is also 0.499 moles.

Calculating the Volume of Hydrogen Gas Produced (V)

To calculate the volume of hydrogen gas produced (V), we can use the ideal gas law equation:

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) conditions: - T = 273.15 K - P = 1 atm

Using the ideal gas law equation: V = (nRT) / P = (0.499 moles * 0.0821 L·atm/mol·K * 273.15 K) / 1 atm ≈ 11.20 L

The amount of hydrogen gas produced (m) is approximately 0.499 moles, and the volume of hydrogen gas produced (V) is approximately 11.20 liters.