A calcium chloride solution was added to 200 g of a silver nitrate solution until the precipitation ceased
A calcium chloride solution was added to 200 g of a silver nitrate solution until the precipitation ceased. The mass of the resulting precipitate was 28.7 g. Calculate the mass fraction of silver nitrate in the original solution.
Let’s execute the solution:
1. Let’s write the equation in accordance with the condition of the problem:
m = 200 g. m = 28.7 g. W -?
2AgNO3 + CaCl2 = 2AgCl + Ca (NO3) 2 – ion exchange, precipitate of chloride is released
2. Calculation of the molar masses of substances:
M (AgNO3) = 293.8 g / mol;
M (AgCl) = 143.3 g / mol.
3. Calculate the number of moles of silver nitrate, silver chloride:
Y (AgNO3) = m / M = 200 / 293.8 = 0.68 mol;
Y (AgCl) = 0.68 mol since the amount of these substances according to the equation is 1 mol.
4. Find the theoretical mass:
m (AgCl) = Y * M = 0.68 * 143.3 = 97.44 g.
5. Let’s calculate the product yield:
W = m (practical) / m (theoretical) * 100;
W = 28.7 / 97.44 = 29.45%.
Answer: the mass fraction of the silver chloride yield was 29.45%.