Question

Рассчитайте массу негашеной извести, которое образовалось при прокаливании известняка массой 1 кг, содержащего 5 % примесей

Answer 1

СaCO3=CaO+CO2
w(CaCO3) чист=100%-5%=95%
m(CaCO3)чист=m(CaCO3)c примес.*w(CaCO3)=1000 г*0,95=950 г
n(CaCO3)=m/M=950/100=0.95 моль
n(СaO)=n(CaCO3)=0.95 моль
m(CaO)=n*M=0.95*56=53.2 г

Answer 2

Calculating the Mass of Unburned Lime

To calculate the mass of unburned lime formed during the calcination of limestone, we can use the given information about the mass of the limestone and the percentage of impurities.

Given: - Mass of limestone (CaCO3) = 1 kg - Percentage of impurities = 5%

We can use the chemical equation for the calcination of limestone to calculate the mass of unburned lime formed.

The chemical equation for the calcination of limestone is: ``` CaCO3(s) -> CaO(s) + CO2(g) ```

Calculation Steps:

1. Calculate the molar mass of CaCO3. 2. Calculate the moles of CaCO3 in 1 kg of limestone. 3. Determine the moles of CaO formed from the moles of CaCO3. 4. Calculate the mass of CaO formed.

Calculation:

1. Calculate the molar mass of CaCO3: The molar mass of CaCO3 can be calculated as follows: ``` Molar mass of CaCO3 = 40.08 (Ca) + 12.01 (C) + 3(16.00) (3 O) = 100.09 g/mol ```

2. Calculate the moles of CaCO3 in 1 kg of limestone: Using the molar mass of CaCO3, we can calculate the moles of CaCO3 in 1 kg of limestone: ``` Moles of CaCO3 = Mass of limestone / Molar mass of CaCO3 = 1000 g / 100.09 g/mol ≈ 9.99 moles ```

3. Determine the moles of CaO formed from the moles of CaCO3: From the balanced chemical equation, 1 mole of CaCO3 produces 1 mole of CaO. Therefore, the moles of CaO formed will be the same as the moles of CaCO3.

Moles of CaO formed = 9.99 moles

4. Calculate the mass of CaO formed: Using the molar mass of CaO, we can calculate the mass of CaO formed: ``` Molar mass of CaO = 40.08 (Ca) + 16.00 (O) = 56.08 g/mol Mass of CaO formed = Moles of CaO formed * Molar mass of CaO = 9.99 moles * 56.08 g/mol ≈ 560.55 g ```

Result:

The mass of unburned lime (CaO) formed during the calcination of 1 kg of limestone containing 5% impurities is approximately 560.55 grams.