Answer:
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.
