Question

Срочно нужна ваша помощь.Помогите кто чем может, пожалуйста! 45 г кальция содержащего 5% примесей, опустили в воду. Определите объем выделившегося газа?

Answer 1

Ca+2H2O=Ca(OH)2 + H2
1. Найдем массу кальция без примесей:
100-5 = 95%
m(Ca) = 95 x 45 / 100 = 42.75 г
2. Найдем количество кальция, зная его молярную массу и массу:
n(Ca) = m / M = 42.75 / 40 = 1.07 моль
3. По уравнению реакции видно, что n(Ca) / n(H2) = 1 / 1 (количества равны), значит
n(H2) = n(Ca) = 1.07 моль
4. Зная количество водорода, находим его объем:
V(H2) = Vm x n = 22.4 x 1.07 = 23.97 л

Answer 2

Calculating the Volume of Gas Released

To calculate the volume of gas released when 45g of calcium containing 5% impurities is immersed in water, we can use the chemical reaction between calcium and water to determine the volume of gas produced.

The balanced chemical equation for the reaction between calcium and water is:

Ca + 2H2O → Ca(OH)2 + H2

According to the stoichiometry of the reaction, 1 mole of calcium reacts with 2 moles of water to produce 1 mole of hydrogen gas.

Given that the molar mass of calcium is approximately 40.08 g/mol and the molar mass of hydrogen is approximately 2.016 g/mol, we can calculate the moles of calcium and then use the stoichiometry of the reaction to determine the volume of gas produced.

Calculation Steps

1. Calculate the moles of calcium. 2. Use the stoichiometry of the reaction to determine the moles of hydrogen gas produced. 3. Convert the moles of hydrogen gas to volume using the ideal gas law.

Calculation

1. Moles of calcium: - Molar mass of calcium = 40.08 g/mol - Mass of calcium = 45g - Moles of calcium = Mass / Molar mass = 45g / 40.08 g/mol = 1.122 moles

2. Moles of hydrogen gas produced: - According to the balanced chemical equation, 1 mole of calcium produces 1 mole of hydrogen gas. - Moles of hydrogen gas = 1.122 moles

3. Volume of gas produced: - Using the ideal gas law, PV = nRT, where P is pressure, V is volume, n is moles, R is the gas constant, and T is temperature. - Assuming standard temperature and pressure (STP), where T = 273.15 K and P = 1 atm, and using R = 0.0821 L·atm/(K·mol): - V = nRT/P = (1.122 moles)(0.0821 L·atm/(K·mol))(273.15 K) / 1 atm ≈ 24.6 L

Conclusion

The volume of gas released when 45g of calcium containing 5% impurities is immersed in water is approximately 24.6 liters.