Calculate the mass of copper oxide and the mass of water formed during the decomposition of 0.5 mol of copper hydroxide.

Let’s implement the solution:

According to the condition of the problem, we will write the data:
Y = 0.5 mol X g -? X g -?

Cu (OH) 2 = CuO + H2O – decomposition, copper oxide, water is released;

Calculations by formulas:
M (CuO) = 79.5 g / mol;

M (H2O) = 18 g / mol;

Y Cu (OH) 2 = 0.5 mol;

Y (CuO) = 0.5 mol, Y (H2O) = 0.5 mol since the amount of substances is 1 mol.

We find the masses of products:
m (CuO) = Y * M = 0.5 * 79.5 = 39.75 g;

m (H2O) = Y * M = 0.5 * 18 = 9 g

Answer: obtained copper oxide weighing 39.75 g, water – 9 g