Question

Какой объем водорода (н.у) выделится при взаимодействии магния количеством вещества 0,75 моль с избытком уксусной кислоты?

Answer 1

  дано: n(Mg) = 0,75 моль
найти: V(H2)
Решение:
Mg +2CH3COOH = (CH3COO)2Mg + H2
n(H2) : n(Mg) = 1 :1 следовательно:
n(Н2)= 0,75 моль
V(H2) = 22,4 * 0,75=16,8л
Ответ:16,8 литра

Answer 2

Calculating the Volume of Hydrogen Gas Produced

To calculate the volume of hydrogen gas produced when magnesium reacts with an excess of acetic acid, we can use the balanced chemical equation for the reaction and the ideal gas law.

The balanced chemical equation for the reaction between magnesium and acetic acid is as follows: Mg + 2CH3COOH → Mg(CH3COO)2 + H2

Given: - Moles of magnesium (n) = 0.75 mol - Excess acetic acid

Using the ideal gas law equation: PV = nRT

Where: - P = pressure - V = volume - n = number of moles - R = ideal gas constant - T = temperature

We can assume standard temperature and pressure (STP) for the reaction, where T = 273 K and P = 1 atm.

Calculation Steps

1. Calculate the moles of hydrogen gas produced using the stoichiometry of the reaction. 2. Use the ideal gas law to calculate the volume of hydrogen gas produced.

Calculation

1. Moles of hydrogen gas produced: - From the balanced chemical equation, 1 mole of magnesium produces 1 mole of hydrogen gas. - Therefore, 0.75 moles of magnesium will produce 0.75 moles of hydrogen gas.

2. Volume of hydrogen gas produced: - Using the ideal gas law equation: PV = nRT - At STP, T = 273 K and P = 1 atm. - Substituting the values, we get: V = (0.75 mol) * (0.0821 L*atm/mol*K) * (273 K) / (1 atm) - Solving for V, we get: V ≈ 16.6 L

Therefore, the volume of hydrogen gas produced when 0.75 moles of magnesium react with an excess of acetic acid is approximately 16.6 liters. [[1 #]], [[2 #]]