Question

The percentage of sulfur in coal can be determined by burning a sample of the solid and passing the resulting sulfur dioxide gas into a solution of hydrogen peroxide, which oxidizes it to sulfuric acid, and then titrating the acid. Calculate the mass percent of sulfur in a sample if the gas from an 8.05-g sample required 44.1 mL of 0.114 M NaOH in the titration of the diprotic acid.

Answer 1

The mass percent of sulfur is 1.00 %

Step-by-step explanation:

Step 1: Data given

Mass of the sample = 8.05 grams

Volume of a 0.114 M NaOH = 44.1 mL = 0.0441 L

Molar mass of sulfur = 32.065 g/mol

Step 2: The balanced equation

H2SO4 + 2NaOH → Na2SO4 + 2H2O

Step 3: Calculate moles NaOH

Moles NaOH = molarity * volume

Moles NaOH = 0.114 M * 0.0441 L

Moles NaOH = 0.00503 moles

Step 4: Calculate moles H2SO4

For 1 mol H2SO4 we need 2 moles NaOH

For 0.00503 moles NaOH we need 0.00503/2 = 0.002515 moles

Step 5: Calculate moles sulfur

For 1 mol H2SO4 we get 1 mol sulfur

For 0.002515 moles H2SO4 we have 0.002515 moles Sulfur

Step 6: Calculate mass of Sulfur

Mass sulfur = moles sulfur * molar mass sulfur

Mass sulfur = 0.002515 moles * 32.065 g/mol

Mass sulfur = 0.0806 grams

Step 7: Calculate % of sulfur

% sulfur = (0.0806 / 8.05)*100%

% sulfur = 1.00%

The mass percent of sulfur is 1.00 %

Answer 2

The mass percent of sulfur is 9.98%.

Step-by-step explanation:

The percentage of sulfur in coal can be determined by burning a sample of the solid and passing the resulting sulfur dioxide gas into a solution of hydrogen peroxide, which oxidizes it to sulfuric acid, and then titrating the acid.

The corresponding balanced equations are:

S + O₂ ⇄ SO₂ [1]

SO₂ + H₂O₂ ⇄ H₂SO₄ [2]

H₂SO₄ + 2 NaOH ⇄ Na₂SO₄ + 2 H₂O [3]

First, we have to calculate how many moles of NaOH we have.

n = 0.114 mol/L × 0.441 L = 0.0502 mol

Then, we will use the following relations:

• According to [3], 2 moles of NaOH react with 1 mole of H₂SO₄
• According to [2], 1 mole of H₂SO₄ is produced per mole of SO₂
• According to [1], 1 mole of SO₂ is produced per mole of S
• The molar mass of S is 32.0 g/mol

Then, for 0.0502 moles of NaOH:

`0.0502molNaOH.(1molH_(2)SO_(4))/(2molNaOH) .(1molSO_(2))/(1molH_(2)SO_(4)) .(1molS)/(1molSO_(2)) .(32.0gS)/(1molS) =0.803gS`

If there are 0.803 g of S in an 8.05 g sample, the mass percent of sulfur is:

`(0.803g)/(8.05g) .100 \%= 9.98\%`