To a solution containing zinc chloride weighing 20.4 grams, a solution containing silver nitrate weighing 25.2 grams

| December 3, 2020 | education

To a solution containing zinc chloride weighing 20.4 grams, a solution containing silver nitrate weighing 25.2 grams was added. What substances are in the solution after the reactions? Calculate the amount and mass of each of these substances. Calculate the mass of the precipitate formed

ZnCl2 + 2AgNO3 = 2AgCl + Zn (NO3) 2
n (ZnCl2) = m \ M = 20.4 \ 136 = 0.15 mol
n (AgNO3) = m \ M = 25.2 \ 170 = 0.15 mol
AgNO3 is in short supply, we calculate on it
The solution after the reaction will contain: excess ZnCl2 and the formed Zn (NO3) 2
n (AgCl) at ur-th p-i = 0.15 mol m precipitate (AgCl) = nM = 0.15 * 143.5 = 21.525 g
n (Zn (NO3) 2) at ur-th p-u = 0.075 mol m (Zn (NO3) 2) = nM = 0.075 * 189 = 14.175 g
excess ZnCl2 = 0.15-0.075 (n proreag) = 0.075 mol
m (ZnCl2) = nM = 0.075 * 136 = 10.2 g
Answer: m (ZnCl2) = 10.2 g
m (AgCl) = 21.525 g
m (Zn (NO3) 2) = 14.175 g



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