If 240.0 g NaOH reacts, how many grams of Cu(OH)2 should precipitate out in the reaction? Show work

292.7 g

Explanation:

Note: the reaction is missing but I looked it up on the web and found it.

Step 1: Write the balanced equation

CuSO₄(aq) + 2 NaOH(aq) → Cu(OH)₂(s) + Na₂SO₄(aq)

Step 2: Calculate the moles corresponding to 240.0 g of NaOH

The molar mass of NaOH is 40.00 g/mol.

240.0 g × 1 mol/40.00 g = 6.000 mol

Step 3: Calculate the moles of Cu(OH)₂ formed from 6.000 moles of NaOH

The molar ratio of NaOH to Cu(OH)₂ is 2:1. The moles of Cu(OH)₂ formed are 1/2 × 6.000 mol = 3.000 mol.

Step 4: Calculate the mass corresponding to 3.000 moles of Cu(OH)₂

The molar mass of Cu(OH)₂ is 97.56 g/mol.

3.000 mol × 97.56 g/mol = 292.7 g