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Sulfuric acid is produced by first burning sulfur to produce sulfur trioxide gas

2S(s) + 3O2(g) → 2SO3(g)

then dissolving the sulfur trioxide gas in water

SO3(g) + H2O(l) → H2SO4(l)

Calculate the mass of sulfuric acid produced if 1.25 g of sulfur is reacted as indicated in the above equations.

User Eselfar
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1 Answer

2 votes

Answer:

3.82 g.

Step-by-step explanation:

To calculate the mass of sulfuric acid produced from 1.25 g of sulfur, we'll have to balance the chemical equations and use stoichiometry.

Starting with the first reaction:

2S(s) + 3O2(g) → 2SO3(g)

Since 1.25 g of sulfur is reacted, the number of moles of sulfur can be calculated as:

moles = mass / molar mass = 1.25 g / 32 g/mol = 0.03906 mol

Next, using the mole ratio from the balanced chemical equation, we can find the number of moles of sulfur trioxide produced:

2 moles of sulfur produce 2 moles of sulfur trioxide, so:

moles of SO3 = moles of S x (moles of SO3 / moles of S) = 0.03906 mol x (2 mol / 2 mol) = 0.03906 mol

Finally, we move on to the second reaction, the dissolution of sulfur trioxide in water:

SO3(g) + H2O(l) → H2SO4(l)

Using the mole ratio from this balanced equation, we can find the number of moles of sulfuric acid produced:

1 mole of sulfur trioxide reacts with 1 mole of water to produce 1 mole of sulfuric acid, so:

moles of H2SO4 = moles of SO3 x (moles of H2SO4 / moles of SO3) = 0.03906 mol x (1 mol / 1 mol) = 0.03906 mol

The mass of sulfuric acid produced can be calculated using the moles and the molar mass of sulfuric acid:

mass = moles x molar mass = 0.03906 mol x 98 g/mol = 3.82 g

Therefore, if 1.25 g of sulfur is reacted, the mass of sulfuric acid produced is approximately 3.82 g.

User Greg Witczak
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