What is the mass of the precipitate formed by the interaction of 20 g of a 15% sodium

| April 29, 2021 | education

What is the mass of the precipitate formed by the interaction of 20 g of a 15% sodium chloride solution with an excess of silver nitrate solution?

Find the mass of sodium chloride in the solution.

W = m (substance): m (solution) × 100%, hence

m (substance) = (m (solution) × W): 100%.

m (substance) = (20 g × 15%): 100% = 3 g.

Find the amount of magnesium nitrate NaCl by the formula:

n = m: M.

M (NaCl) = 58.5 g / mol.

n = 3 g: 58.5 g / mol = 0.05 mol.

Let’s compose the reaction equation, find the quantitative ratios of substances.

NaCl + AgNO3 = AgCl ↓ + NaNO3.

According to the reaction equation, 1 mol of NaCl accounts for 1 mol of AgCl. Substances are in quantitative ratios 1: 1.

The amount of substance will be equal.

n (AgCl) = n (NaCl) = 0.05 mol.

Let us find the mass of AgCl.

M (AgCl) = 143.5 g / mol.

m = n × M.

m = 143.5 g / mol × 0.05 mol = 7.175 g.

Answer: 7.175 g.



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