What mass of salt is formed by the interaction of higher calcium oxide weighing 6.0 g

| August 7, 2021 | education

What mass of salt is formed by the interaction of higher calcium oxide weighing 6.0 g with higher carbon monoxide with a volume of 22.4 dm3

Let’s implement the solution:
1. In accordance with the condition of the problem, we write the equation:
CaO + CO2 = CaCO3 – compounds, calcium carbonate was formed;
1 mol 1 mol 1 mol.
2. Calculations according to the formulas of substances:
M (CaO) = 56 g / mol;
M (CO2) = 44 g / mol;
M (CaCO3) = 100 g / mol;
V (CO2) = 22.4 dm3 = 22.4 l.
3. Determine the amount of the original substance:
Y (CaO) = m / M = 6/56 = 0.1 mol (deficient substance);
Y (CO2) = 1 mol, since one mol takes up a volume of 22.4 liters. according to Avogadro’s law. (substance in excess);
Calculations are made taking into account the deficient substance.
Y (CaCO3) = 0.1 mol.
4. Find the mass of the product:
m (CaCO3) = Y * M = 0.1 * 100 = 10 g.
Answer: Received calcium carbonate weighing 10 g.



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