To test the purity of sodium bicarbonate, you dissolve a 3.50g sample in water and add sulfuric acid. if 1.04g of carbon dioxide forms, what is the percent purity of the sodium bicarbonate?

Answer is: the percent purity of the sodium bicarbonate is 56.83 %.

1. Chemical reaction: 2NaHCO₃ + H₂SO₄ → 2CO₂ + 2H₂O + Na₂SO₄.

2. m(NaHCO₃) = 3.50 g

n(NaHCO₃) = m(NaHCO₃) ÷ M(NaHCO₃).

n(NaHCO₃) = 3.50 g ÷ 84 g/mol.

n(NaHCO₃) = 0.042 mol.

3. From chemical reaction: n(NaHCO₃) : n(CO₂) = 1 : 1.

n(CO₂) = 0.042 mol.

m(CO₂) = 0.042 mol · 44 g/mol.

m(CO₂) = 1.83 g.

4. the percent purity = 1.04 g/1.83 g  ·100%.

the percent purity = 56.8 %.

it's filtration, but this is an assumption so my apologies if i got this