What masses of salt and water are formed when copper (II) oxide is dissolved in 300 g of 9.8% sulfuric acid solution?

Let’s write the reaction equation:
CuO + H2SO4 = CuSo4 + H2O
Find the mass of the acid:
w (in-va) = (m (in-va) / m (solution)) * 100%
m (in-va) = w (in-va) * m (solution) / 100% = 9.8 * 300/100 = 29.4 g.
It can be seen from the reaction equation that:
ν (H2SO4) = ν (CuSo4) = ν (H2O)
m (H2SO4) / M (H2SO4) = m (CuSo4) / M (CuSo4)
m (H2SO4) / M (H2SO4) = m (H2O) / M (H2O)
Let’s define the molar masses:
M (H2SO4) 1 * 2 + 32 + 16 * 4 = 98 g / mol
M (CuSo4) = 64 + 32 + 16 * 4 = 160 g / mol
M (H2O) = 1 * 2 + 16 = 18 g / mol
Determine the masses of salt and water:
m (CuSo4) = m (H2SO4) * M (CuSo4) / M (H2SO4) = 29.4 * 160/98 = 48 g.
m (H2O) = m (H2SO4) * M (H2O) / M (H2SO4) = 29.4 * 18/98 = 5.4 g.
Answer: the mass of salt is 48 g, the mass of water is 5.4 g.