Calculate the mass of the precipitate formed by the interaction of 100 g of 10% silver nitrate solution and hydrochloric acid solution.
In the reaction of silver nitrate with hydrochloric acid, silver chloride precipitates.
AgNO3 + HCl = AgCl + HNO3.
Determine the mass of silver nitrate in solution.
m (AgNO3) = W (AgNO3) • m (solution) / 100% = 10% • 100 g / 100% = 10 g.
The number of moles of nitrate is equal to the number of moles of silver chloride.
n (AgNO3) = n (AgCl) = m / Mr = 10 g / 168.87 g / mol = 0.06 mol.
The mass of the dropped out salt.
m (AgCl) = n • Mr = 0.06 mol • 143.32 g / mol = 8.44 g.
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