Question

A 26.08 g mixture of zinc and sodium is reacted with a stoichiometric amount of sulfuric acid. The reaction mixture is then reacted with 132 mL of 3.80 M barium chloride to produce the maximum possible amount of barium sulfate. Determine the percent sodium by mass in the original mixture.

Answer 1

Answer:

Molar percent of sodium in original mixture is 88,50%

Explanation:

The last reaction is:

BaCl₂ + Na₂SO₄ → BaSO₄ + 2 NaCl

The moles of BaCl₂ are:

0,132L × 3,80M = 0,502 moles of BaCl₂

As the amount of BaCl₂ is the maximum possible to produce BaSO₄, the moles of BaCl₂ must be the same than moles of Na₂SO₄.

0,502 moles of BaCl₂ ≡ 0,502 moles of Na₂SO₄

These moles of Na₂SO₄ comes from:

2 Na + H₂SO₄ → Na₂SO₄ + H₂

As the reaction is in stoichiometric amounts, moles of Na are twice the moles of Na₂SO₄

0,502 moles of Na₂SO₄ ×$\frac{2molesNa}{1moleNa_{2}SO_{4}}$× 22,99 g/mole = 23,08 g of Na

Molar percent of sodium in original mixture is:

$\frac{23,08g}{26,08g}*100$ = 88,50%

I hope it helps