Calculate the mass of the salt formed by the interaction of sulfuric acid weighing 9.8 g with a solution containing calcium hydroxide weighing 14.8 g
Given: m (H2SO4) = 9.8 g m (Ca (OH) 2) = 14.8 g Find: m (salt form) -?
1) write the reaction equation: H2SO4 + Ca (OH) 2 => CaSO4 ↓ + 2H2O 2) find the amount of substance H2SO4 and Ca (OH) 2: n (H2SO4) = m (H2SO4) / Mr (H2SO4) = 9.8 / 98 = 0.1 mol n (Ca (OH) 2) = m (Ca (OH) 2) / Mr (Ca (OH) 2) = 14.8 / 74 = 0.2 mol 3) compare the amount of substance H2SO4 and Ca (OH) 2: according to the reaction equation, the same amount of H2SO4 and Ca (OH) 2 interact, but according to the condition Ca (OH) 2 is in excess (0.2 mol). Therefore, the calculations will be carried out for a substance that completely reacts, that is, H2SO4 (0.1 mol) 4) find the amount of CaSO4 (taking into account the reaction equation): n (CaSO4) = n (H2SO4) = 0.1 mol 5) find mass of CaSO4 substance: m (CaSO4) = n (CaSO4) * Mr (CaSO4) = 0.1 * 136 = 13.6 g
Answer: the mass of the formed salt (CaSO4) is 13.6 g.
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