The mass of the precipitate formed as a result of the interaction of a solution of silver nitrate containing 8.5 g
| December 11, 2020 | education
The mass of the precipitate formed as a result of the interaction of a solution of silver nitrate containing 8.5 g of salt with a solution of hydrochloric acid is …
Given:
m (AgNO3) = 8.5 g
Find: m (AgCL)
Decision:
1) Let’s write the reaction equation:
AgNO3 + HCl = AgCl + HNO3
2) Calculate the amount of AgNO3:
n (AgNO3) = m (AgNO3) / M (AgNO3) = 8.5 / 170 = 0.05 mol
3) According to the reaction equation n (AgNO3) = n (AgCl) = 0.05 mol
4) Calculate the mass of the sediment:
m (AgCl) = M (AgCl) * n (AgCl) = 143.5 * 0.05 = 7.175 mol
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