S(s) + 6HNO3(aq) → H2SO4(aq) + 6NO2(g) + 2H2O(l)

Question

The equation shows the reaction of sulfur with hot concentrated nitric acid:

        S(s) + 6HNO₃(aq) → H₂SO₄(aq) + 6NO₂(g) + 2H₂O(l)

Calculate the mass of nitrogen dioxide made, in grams, when 2.0 g of sulfur reacts completely with nitric acid.

Give your answer to two significant figures.

Answer 1

1. Map out the problem:

Notice that you'll first need to convert the 2g of sulfur into moles. You then will need to convert the moles of sulfur into moles of NO₂ and lastly you need to convert those moles of NO₂ into grams of NO₂.

2. Convert 2g of sulfur into moles of sulfur. To do that, use the periodic table and look up the molar mass of sulfur. Mine says sulfur is 32.07 g/mol. So 2g divided by 32.07g/mol = 0.06236 moles.

3. Convert 0.06236 moles of sulfur into moles of NO₂. To do that, use the mole-to-mole ratio in the balanced chemical reaction equation. Notice that for 1 mole of sulfur, 6 moles of NO₂ is produced. So to calculate the number of moles of NO₂, multiply 0.06236 by 6: (0.06236)(6) = 0.3742 moles of NO₂.

4. Convert moles of NO₂ into grams of NO₂. To do that, calculate the molar mass of NO₂: From the periodic table: 14.01g/mol (nitrogen) + 16.00(2) g/mol (oxygen) = 14.01 + 32.00 = 46.01 g/mol. Now to calculate the grams of NO2: 0.3742 moles of NO₂ multiplied by 46.01 g/mol = 17 grams of NO₂