Question

При обжиге карбоната кальция получили 5.6 литра углекислого газа при нормальных условиях. Найти массу оксида кальция, полученной при обжиге.

Answer 1

 Дано:

V (CO2) =  5,6 л

Найти:

m(CaO)

Решение:

                  x г         5,6 л

CaCO3 = CaO + CO2↑

                 56 г       22,4 л          

x  г --- 5,6 л

56г --- 22,4 л

х = 56 г ∙ 5,6 л\22,4 л = 14 г

ответ: m(CаO) = 14 г

Answer 2

Calculation of Mass of Calcium Oxide Produced

To calculate the mass of calcium oxide produced during the combustion of calcium carbonate, we need to use the balanced chemical equation for the reaction. The balanced equation for the combustion of calcium carbonate is:

CaCO3(s) → CaO(s) + CO2(g)

From the equation, we can see that one mole of calcium carbonate (CaCO3) produces one mole of calcium oxide (CaO) and one mole of carbon dioxide (CO2).

Given that 5.6 liters of carbon dioxide gas is produced at standard conditions (STP), we can use the ideal gas law to convert the volume of gas to moles. At STP, one mole of any ideal gas occupies 22.4 liters.

Using the equation:

moles = volume (liters) / molar volume (liters/mole)

we can calculate the number of moles of carbon dioxide gas produced:

moles of CO2 = 5.6 liters / 22.4 liters/mole = 0.25 moles

Since the balanced equation tells us that one mole of calcium carbonate produces one mole of calcium oxide, the number of moles of calcium oxide produced is also 0.25 moles.

To find the mass of calcium oxide produced, we need to know the molar mass of calcium oxide. The molar mass of calcium oxide (CaO) is the sum of the atomic masses of calcium (Ca) and oxygen (O).

Using the atomic masses from the periodic table:

molar mass of CaO = atomic mass of Ca + atomic mass of O = 40.08 g/mol + 16.00 g/mol = 56.08 g/mol

Finally, we can calculate the mass of calcium oxide produced:

mass of CaO = moles of CaO * molar mass of CaO = 0.25 moles * 56.08 g/mol = 14.02 grams

Therefore, the mass of calcium oxide produced during the combustion of calcium carbonate is approximately 14.02 grams.