How many grams of copper (2) oxide, carbon dioxide and water are formed during the decomposition

How many grams of copper (2) oxide, carbon dioxide and water are formed during the decomposition of 444 g of malachite CU2 (OH) 2 CO3?

Let’s execute the solution:

In accordance with the condition of the problem, we write:

(CuOH) 2CO3 = 2CuO + CO2 + H2O – decomposition of malachite (basic copper carbonate), copper oxide (2), carbon monoxide (2), water are released;

Calculations:
M (CuOH) 2CO3 = 221 g / mol;

M (CuO) = 79.5 g / mol;

M (CO2) = 44 g / mol;

M (H2O) = 18 g / mol.

Y (CuOH2) CO3 = m / M = 444/221 = 2 mol;

Y (CO2) = 2 mol; Y (H2O) = 2 mol since the amount of these substances is equal to 1 mol.

Proportion:
2 mol of malachite – X mol (CuO);

-1 mol -2 mol from here, X mol (CuO) = 2 * 2/1 = 4 mol.

We find the masses of products:
m (CuO) = Y * M = 4 * 79.5 = 318 g;

m (CO2) = Y * M = 2 * 44 = 88 g;

m (H2O) = 2 * 18 = 36 g

Answer: the mass of copper oxide was 318 g, carbon monoxide – 88 g, water – 36 g