Carbon monoxide (4) reacted with calcium oxide weighing 504 grams. Find the mass of salt formed.

Decision:
1. Let’s compose the reaction equation:
CO2 + CaO = CaCO3 – compound reaction, calcium carbonate salt is obtained;
2. Let’s make calculations using the formulas:
M (CaO) = 40 + 16 = 56 g / mol;
M (CaCO3) = 40 +12 +16 * 3 = 100 g / mol;
3. Determine the number of moles of CaO, if the mass is known:
Y (CaO) = m / M = 504/56 = 9 mol;
4. Let’s make the proportion according to the equation:
9 mol (CaO) – X mol (CaCO3);
– 1 mol -1 mol from here, X mol (CaCO3) = 9 * 1/1 = 9 mol;
5. Let’s calculate the mass of CaCO3 by the formula:
M (CaCO3) = Y * M;
m (CaCO3) = 9 * 100 = 900 g;
Answer: in the course of the reaction, calcium carbonate weighing 900 g was formed.