How many grams of sediment is formed by the interaction of 16.4 grams of “sodium phosphate” and 70 grams of “silver nitrate”.
August 29, 2021 | education|
To solve, we compose the equation of the process:
Na3PO4 + 2AgNO3 = 3NaNO3 + Ag3PO4 – ion exchange, a precipitate of silver phosphate is formed;
Let’s make calculations using the formulas of substances:
M (Na3PO4) = 163.6 g / mol;
M (AgNO3) = 169.8 g / mol;
M (Ag3PO4) = 418.3 g / mol.
Let’s determine the amount of starting substances:
Y (Na3PO4) = m / M = 16.4 / 163.6 = 0.1 mol (deficient substance);
Y (AgNO3) = 0.1 mol since the amount of these substances is 1 mol.
Y (AgNO3) = m / M = 70 / 169.8 = 0.4 mol (substance in excess).
Calculations are made for the substance in deficiency.
Find the mass of the product:
m (Ag3PO4) = Y * M = 0.1 * 418.3 = 41.83 g
Answer: The mass of silver phosphate is 41.83 g
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