How many grams of precipitate is formed when copper (2) sulfate interacts with sodium hydroxide weighing 120 grams?

In order to find how many grams of precipitate is formed during the interaction of copper sulfate (2) with sodium hydroxide weighing 120 grams, we first of all draw up an equation for the reaction of interaction.

CuSO4 + NaOH = Cu (OH) 2 + Na2SO4.

Let’s arrange the coefficients in the reaction equation:

CuSO4 + 2NaOH = Cu (OH) 2 + Na2SO4.

We will start the solution by writing down the mass of 1 mole of substances:

M (NaOH) = 40 g / mol;

M (Cu (OH) 2) = 98 g / mol.

To calculate the mass of Cu (OH) 2, we proceed as follows:

m (Cu (OH) 2) = (120 * 98) / 80 = 147 grams.