Question

CaCO3-CO2+CaO маса CaCO3 120г знайти масу CO якшо вихід її становить 85%

Answer 1

 120 г                x г
CaCO3=CaO+CO2
100 г/моль     44 г/моль

mтеор(CO2)=m(CaCO3)*M(CO2)/M(CaCO3)=120 г*44 г/моль/100 г/моль=52,8 г.
mпрак.=mтеор*Wпрак/100 %=52,8 г*85%/100%=44,88 г
Ответ: 44,88 г или примерно 45 г.

Answer 2

Calculation of the Mass of CO Produced from the Decomposition of CaCO3

To calculate the mass of CO produced from the decomposition of CaCO3, we need to consider the balanced chemical equation for the reaction:

CaCO3 -> CO2 + CaO

According to the equation, one mole of CaCO3 decomposes to produce one mole of CO2 and one mole of CaO. The molar mass of CaCO3 is 100.09 g/mol, which means that 100.09 grams of CaCO3 will produce 44.01 grams of CO2.

Given that the mass of CaCO3 is 120 grams and the yield of the reaction is 85%, we can calculate the mass of CO produced as follows:

1. Calculate the moles of CaCO3: - Moles of CaCO3 = Mass of CaCO3 / Molar mass of CaCO3 - Moles of CaCO3 = 120 g / 100.09 g/mol = 1.199 moles

2. Calculate the moles of CO2 produced: - Moles of CO2 = Moles of CaCO3 (since the reaction is 1:1) - Moles of CO2 = 1.199 moles

3. Calculate the mass of CO2 produced: - Mass of CO2 = Moles of CO2 * Molar mass of CO2 - Mass of CO2 = 1.199 moles * 44.01 g/mol = 52.79 grams

4. Calculate the mass of CO produced: - Mass of CO = Mass of CO2 * Yield - Mass of CO = 52.79 grams * 0.85 = 44.87 grams

Therefore, the mass of CO produced from the decomposition of 120 grams of CaCO3 with a yield of 85% is approximately 44.87 grams.