When 75g of sodium sulfite and 80.0g of hydrogen bromide are reacted, the products formed are sodium bromide and hydrogen sulfite. Calculate the mass of sodium bromide produced.

Answer 1

Step 1: Write a reaction Equation (don't worry about balancing it but make sure the compounds are correctly written)

Sodium Sulfite: Sodium is Na and the ion is a +1 while sulfite is SO₃ and is a -2 so it will take 2 sodium ions making sodium sulfite Na₂SO₃

Hydrogen Bromide: Hydrogen is +1 and the bromine ion (bromide) is -1 so hydrogen bromide is HBr

Sodium bromide is NaBr

Hydrogen sulfide is H₂SO₃

So an unbalanced chemical reaction equation is:

Na₂SO₃ + HBr → NaBr + H₂SO₃

Step 2: Balance the chemical reaction equation:

Na₂SO₃ + 2HBr → 2NaBr + H₂SO₃

Step 3: Convert grams to moles:

First calculate molar masses for the compounds:

The molar mass of Na₂SO₃ is:

Na: (22.990)(2) = 45.98

S: 32.066

O: (15.999)(3) = 47.997

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Na₂SO₃: 126.043 g/mol

The molar mass of HBr is:

H: 1.008

Br: 79.904

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HBr: 80.912 g/mol

The molar mass of NaBr is:

Na: 22.990

Br: 79.904

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NaBr: 102.894 g/mol

The molar mass of H₂SO₃ is:

H: (1.008)(2) = 2.016

S: 32.066

O: (15.999)(3) = 47.997

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H₂SO₃: 82.079 g/mol

75 g Na₂SO₃ / 126.043 g/mol = 0,595 moles Na₂SO₃

80 g HBr / 80.912 g/mol = 0.9887 moles HBr

Step 4: Determine the limiting reactant.

The chemical reaction equation Na₂SO₃ + 2HBr → 2NaBr + H₂SO₃ say I need 2 moles of HBr for every 1 mole of Na₂SO₃ but we don't have twice as much HBr so HBr is the limiting reactant (only 0.9887/2, or 0.494 moles, of the Na₂SO₃ will be consumed leaving 0.1 moles of Na₂SO₃ unused).

Step 5: Use the molar ratios to determine the moles of product produced

Since HBr is the limiting reactant, we only need to worry about that in determining the amount of the products produced. The chemical reaction equation Na₂SO₃ + 2HBr → 2NaBr + H₂SO₃ says that for every 2 moles of HBr used, 2 moles of NaBr are produced so for the 0.9887 moles HBr used, there will be 0.9887 moles NaBr produced.

Step 6: Calculate the mass of the product produced.

0.9887 moles NaBr x 102.894 g/mol = 101 73 g NaBr produced. There are only 2 sig figs in the sodium sulfide (75 g) however, that was not used in the calculation (it was only used in determining the limiting reactant) so I'm going to go with 3 sig figs of the HBr given (80.0 g) meaning my answer should have 3 sig figs or be 102 g NaBr produced