Calculate the mass of the precipitate formed by the interaction of a solution of barium chloride

Calculate the mass of the precipitate formed by the interaction of a solution of barium chloride with a solution of aluminum sulfate containing 68.4 grams of salt.

Given:

m (Al2 (SO4) 3) = 68.4 g

Find:

m (draft) -?

Solution:

Al2 (SO4) 3 + 3BaCl2 = 3BaSO4 + 2AlCl3, – we solve the problem based on the composed reaction equation:

1) Find the amount of aluminum sulfate that has reacted:

n (Al2 (SO4) 3) = m: M = 68.4 g: 342 g / mol = 0.2 mol.

2) We compose a logical expression:

If 1 mol of Al2 (SO4) 3 gives 3 mol of BaSO4,

then 0.2 mol of Al2 (SO4) 3 will give x mol of BaSO4,

then x = 0.6 mol.

3) Find the mass of barium sulfate precipitated during the reaction:

m (BaSO4) = n * M = 0.6 mol * 233 g / mol = 139.8 g.

Answer: m (BaSO4) = 139.8 g.