What is the mass of the precipitate formed when the solutions containing 35 grams of potassium chloride and 50 grams of silver nitrate are poured?
1. We make a chemical equation, not forgetting to put down the coefficients:
KCl + AgNO3 = AgCl + KNO3.
2. Find the amount of substance KCl:
n (KCl) = m / M = 35 / 74.5 = 0.469 mol.
3. Let’s find the amount of substance AgNO3:
n (AgNO3) = m / M = 50/170 = 0.294 mol.
4. We see that:
n (AgNO3) <n (KCl); Therefore, further calculations will be carried out using AgCl (silver nitrate).
5. Find the amount of AgCl:
n (AgCl) = n (AgNO3) = 0.294 mol.
6. Find the mass of AgCl:
m (AgCl) = n * M = 0.294 * 143.5 = 42.189g.
Answer: m (AgCl) = 42.189g.
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