Sulfur oxide (V1) was treated with 140 g of 20% potassium hydroxide solution. Calculate the mass of the salt formed

Sulfur oxide (V1) was treated with 140 g of 20% potassium hydroxide solution. Calculate the mass of the salt formed if the practical reaction yield is 95%.

Given:
m solution (KOH) = 140 g
w (KOH) = 20% = 0.2
n (K2SO4) = 95% = 0.95
To find:
m (K2SO4)
Decision:
2KOH + SO3 = K2SO4 + H2O
m in-va (KOH) = w * m solution = 0.2 * 140 g = 28 g
n (KOH) = m / M = 28 g / 56 g / mol = 0.5 mol
n (KOH): n (K2SO4) = 2: 1
n (K2SO4) = 0.5 mol / 2 = 0.25 mol
m theor (K2SO4) = n * M = 0.25 mol * 174 g / mol = 43.5 g
m prak (K2SO4) = n * m theory = 0.95 * 43.5 g = 41.33 g
Answer: 41.33 g