Question

The amount of BaSo4 formed on mixing an aqueous solution of 2.08 gm of BaCl2 and excess of dilute H2So4 is

Answer 1

First we write a balanced equation of reactants to products. We know BaCl₂ reacts with excess H₂SO₄ to produce BaSO₄ and HCl. The balanced equation is

                                BaCl₂ ₊ H₂SO₄ ⇒  BaSO₄ ₊ 2HCl

Since we know the mass of BaCl₂ is 2.08g, we use the molecular weight of BaCl₂, which is 208.3g/mol to find the moles of BaCl₂ used up in he reaction

$n =\frac{m}{M}$

$=\frac{2.08g}{208.3g/mol}$

$=0.00999mol$

where n is number of moles, m is mass of BaCl₂  ,and M is molecular weight of BaCl₂.

From the balanced equation we know  the ratio BaCl₂: BaSO₄ is 1:1 , hence moles of BaSO₄ will be 0.00999 moles.

We then multiply the number  moles of BaSO₄ with the molecular weight of BaSO₄, which is 98.07g/mol to get the mass of BaSO₄  below:

        $m = nM = (0.00999mol)(98.07g/mol) = 0.989g.$