Calculate the amount and mass of copper hydroxide (||) with which 126 g of nitric acid can be neutralized.

1. Let’s compose the equation of neutralization of nitric acid with copper hydroxide:
2HNO3 + Cu (OH) 2 = Cu (NO3) 2 + 2H2O.
2. Let’s calculate the chemical amount of nitric acid:
n (HNO3) = m (HNO3): M (HNO3);
M (HNO3) = 1 + 14 + 3 * 16 = 63 g / mol;
n (HNO3) = 126: 63 = 2 mol.
3. Determine the amount of copper hydroxide:
n (Cu (OH) 2) = n (HNO3): 2 = 2: 2 = 1 mol.
4. Let’s calculate the mass of hydroxide:
m (Cu (OH) 2) = n (Cu (OH) 2) * M (Cu (OH) 2);
M (Cu (OH) 2) = 64 + 2 * 17 = 98 g / mol;
m (Cu (OH) 2) = 1 * 98 = 98 g.
Answer: 1 mole; 98 g