Sulfur oxide (VI) with a mass of 16 g completely reacted with water. The resulting acid solution was neutralized with a potassium hydroxide solution. Calculate the mass of the salt obtained.
m (SO3) = 16 g
m (salt) -?
1) SO3 + H2O => H2SO4;
H2SO4 + 2KOH => K2SO4 + 2H2O;
2) n (SO3) = m (SO3) / Mr (SO3) = 16/80 = 0.2 mol;
3) n (SO3) = n (H2SO4) = n (K2SO4) = 0.2 mol;
4) m (K2SO4) = n (K2SO4) * Mr (K2SO4) = 0.2 * 174 = 34.8 g.
Answer: The mass of K2SO4 is 34.8.
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