When 80 g of sulfuric acid was neutralized with potassium hydroxide, salt was formed and water was released.

When 80 g of sulfuric acid was neutralized with potassium hydroxide, salt was formed and water was released. Calculate the mass of the salt obtained and the volume of the resulting water.

Given:
m (H2SO4) = 80 g
To find:
m (K2SO4)
V (H2O)
Decision:
H2SO4 + 2KOH = K2SO4 + 2H2O
n (H2SO4) = m / M = 80 g / 98 g / mol = 0.816 mol
n (H2SO4): n (K2SO4) = 1: 1
n (K2SO4) = 0.816 mol
m (K2SO4) = n * M = 0.816 mol * 174 g / mol = 142 g
n (H2SO4): n (H2O) = 1: 2
n (H2O) = 0.816 mol * 2 = 1.632 mol
V (H2O) = n * Vm = 1.632 mol * 22.4 L / mol = 36.56 L
Answer: 142 g; 36.55 l