2. BaO + H2SO4 —> BaSO + H2O
A) How much BaSo4 can be formed from 196.0 g of H2SO4?

B) How many moles of H2O, are formed when 345.0 g of Bao are reacted with excess H2SO4?

A. 466 g of BaSO₄.

B. 2.25 moles of H₂O.

Explanation:

The balanced equation for the reaction is given below:

BaO + H₂SO₄ —> BaSO₄ + H₂O

Next, we shall determine the mass of H₂SO₄ that reacted and the mass of BaSO₄ produced from the balanced equation. This can be obtained as follow:

Molar mass of H₂SO₄ = (2×1) + 32 + (16×4)

= 2 + 32 + 64

= 98 g/mol

Mass of H₂SO₄ from the balanced equation = 1 × 98 = 98 g

Molar mass of BaSO₄ = 137 + 32 + (16×4)

= 137 + 32 + 64

= 233 g/mol

Mass of BaSO₄ from the balanced equation = 1 × 233 = 233 g

SUMMARY:

From the balanced equation above,

98 g of H₂SO₄ reacted to produce 233 g of BaSO₄.

A. Determination of the mass of BaSO₄ produced by the reaction of 196 g of H₂SO₄.

From the balanced equation above,

98 g of H₂SO₄ reacted to produce 233 g of BaSO₄.

Therefore, 196 g of H₂SO₄ will react to produce = (196 × 233) /98 = 466 g of BaSO₄.

Thus, 466 g of BaSO₄ were obtained from the reaction.

B. Determination of the number of mole of H₂O produced by the reaction of 345 g of BaO.

We'll begin by calculating the number of mole in 345 g of BaO. This can be obtained as follow:

Mass of BaO = 345 g

Molar mass of BaO = 137 + 16 = 153 g/mol

Mole of BaO =?

Mole = mass /Molar mass

Mole of BaO = 345 / 153

Mole of BaO = 2.25 moles

Finally, we shall determine the number of mole of H₂O produced from the reaction. This can be obtained as follow:

BaO + H₂SO₄ —> BaSO₄ + H₂O

From the balanced equation above,

1 mole of BaO reacted to produce 1 mole of H₂O.

Therefore, 2.25 moles of BaO will also react to produce 2.25 moles of H₂O.

Thus, 2.25 moles of H₂O were obtained from the reaction.

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