# To 200 g of a solution with a mass fraction of silver nitrate of 8.5% was added 200 g of a solution

**To 200 g of a solution with a mass fraction of silver nitrate of 8.5% was added 200 g of a solution with a mass fraction of potassium chloride of 5%. determine the mass of the sediment formed**

Given:

m (AgNO3) = 200 g

m (KCl) = 200 g

w% (AgNO3) = 8.5%

w% (KCl) = 5%

Find:

m (draft) -?

Solution:

AgNO3 + KCl = AgCl + KNO3, – we solve the problem, relying on the composed reaction equation:

1) Find the masses of silver nitrate and potassium chloride in solutions:

m (AgNO3) = 200 g * 0.085 = 17 g

m (KCl) = 200 g * 0.05 = 10 g

2) Find the amount of silver nitrate and potassium chloride:

n (AgNO3) = m: M = 17 g: 170 g / mol = 0.1 mol

n (KCl) = m: M = 10 g: 74.5 g / mol = 0.134 mol

We start from a lower value to get more accurate calculations. We work with silver nitrate:

3) We compose a logical expression:

if 1 mol of AgNO3 gives 1 mol of AgCl,

then 0.1 mol of AgNO3 will give x mol of AgCl,

then x = 0.1 mol.

4) Find the mass of silver chloride precipitated during the reaction:

m (AgCl) = n * M = 0.1 mol * 143.5 g / mol = 14.35 g.

Answer: m (AgCl) = 14.35 g.