Calculate the mass of the precipitate formed by the interaction of 300 g of silver nitrate with sodium chloride.

Given:

m (AgNO3) = 300 g

Find:

m (draft) -?

Solution:

AgNO3 + NaCl = AgCl + NaNO3, – we solve the problem, relying on the composed reaction equation:

1) Find the amount of silver nitrate that has reacted:

n (AgNO3) = m: M = 300 g: 170 g / mol = 1.76 mol

2) We compose a logical expression:

if 1 mol of AgCl requires 1 mol of AgNO3,

then x mol AgCl will require 1.76 mol AgNO3,

then x = 1.76 mol.

3) Find the mass of silver chloride precipitated as a result of the reaction:

m (AgCl) = n * M = 1.76 mol * 143.5 g / mol = 252.56 g.

Answer: m (AgCl) = 252.56 g.