A solution of silver nitrate was added to hydrochloric acid weighing 365 grams with a mass fraction of 10%.
A solution of silver nitrate was added to hydrochloric acid weighing 365 grams with a mass fraction of 10%. Determine the mass of the resulting residue.
Find the mass of HCl in solution.
W = m (substance): m (solution) × 100%,
m (substance) = (m (solution) × W): 100%.
m (HCl) = (365 g × 10%): 100% = 36.5 g.
Let’s find the amount of HCl by the formula:
n = m: M.
M (HCl) = 36.5 g / mol.
n = 36.5 g: 36.5 g / mol = 1 mol.
Let’s compose the reaction equation, find the quantitative ratios of substances.
HCl + AgNO3 = AgCl ↓ + HNO3.
According to the reaction equation, there is 1 mol of AgCl per 1 mol of HCl. Substances are in quantitative ratios 1: 1.
n (AgCl) = n (HCl) = 1 mol.
Let’s find the mass of AgCl by the formula:
m = n × M,
М (AgCl) = 143.5 g / mol.
m = 1 mol × 143.5 g / mol = 143.5 g.
Answer: 143.5 g.