Calculate the mass of sediment that is formed by the interaction of 24g alcl 80g naoh.

Given:

m (AlCl3) = 24 g

m (NAOH) = 80 g

Find:

m (draft) -?

Solution:

AlCl3 + 3NaOH = 3NaCl + Al (OH) 3, – we solve the problem using the composed reaction equation:

1) Find the amount of aluminum chloride and sodium hydroxide:

n (AlCl3) = m: M = 24 g: 133.5 g / mol = 0.18 mol

n (NaOH) = m: M = 80 g: 40 g / mol = 2 mol

We start from a lower value to get more accurate calculations. We work with AlCl3:

2) We compose a logical expression:

if 1 mol of AlCl3 gives 1 mol of Al (OH) 3,

then 0.18 mol of AlCl3 will give x mol of Al (OH) 3,

then x = 0.18 mol.

3) Find the mass of aluminum hydroxide precipitated during the reaction:

m (Al (OH) 3) = n * M = 0.18 mol * 78 g / mol = 14.04 g.

Answer: m (Al (OH) 3) = 14.04 g.