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Sulfuric acid is the most produced industrial chemical in the world. The chemical process involves many steps – the final step combines oleum (H2S2O7) with water to produce sulfuric acid (H2SO4). For the first batch of acid production, 1.20 kilograms of oleum is reacted with excess water, and 435 grams of sulfuric acid are produced.

• Write the balanced equation for the formation of sulfuric acid from oleum and water.



• Calculate the theoretical yield of sulfuric acid. Work must be shown to earn credit.



• Calculate the percent yield for the sulfuric acid production. Work must be shown to earn credit

1 Answer

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Answer:

Balanced equation: H₂S₂O₇ + H₂O → 2H₂SO₄

Theoretical yield of H₂SO₄: 1322 g

Percent yield for the H₂SO₄ production: 33%

Step-by-step explanation:

First of all, we determine the balanced equation:

H₂S₂O₇ + H₂O → 2H₂SO₄

Stoichiometry says that 1 mol of oleum reacts to 1 mol of water in order to produce 2 moles of sulfuric acid.

1.20 kg = 1200 g

We convert the grams to moles → 1200 g . 1mol/ 178.12g = 6.74 mol

Ratio is 1:2, so if we have 6.74 mol of oleum we can produce the double of moles of acid. (6.74 . 2) = 13.48 moles of acid.

This is the 100 % yield reaction, or it can be called theoretical yield.

We convert the moles to mass → 13.48 mol . 98g /1mol = 1322 g of acid may be produced.

In the problem, they said we produced 435 g of acid, so let's determine the percent yield.

Percent yield of reaction = (Produced yield / Theoretical yield) . 100

Percent yield of reaction = (435 / 1322) . 100 → 33 %

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