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Margarita [4]
3 years ago
8

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?
Chemistry
1 answer:
Andrews [41]3 years ago
8 0

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 %.

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The law of Multiple proportion states that when two elements combine to form more than one compound, the mass of the second element, which combines with a fixed mass of the first element, will always be ratios of small whole numbers.

To illustrate the law of multiple proportion from sulphur and oxygen, we use the given compounds.

The reaction of oxygen with sulphur to form SO₂ and SO₃:

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<em>Notice</em><em>, while mass of </em><em>sulphur</em><em> is constant, mass of </em><em>oxygen</em><em> changed.</em>

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Thus, based on the law of multiple proportion, the mass of the oxygen which combined with a fixed mass of the sulphur is in simple whole number ratio of 2 is to 3 (2:3).

Learn more here: brainly.com/question/15537778

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