Answer:
31.2 g of Ag₂SO₄
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
2AgNO₃(aq) + H₂SO₄ (aq) → Ag₂SO₄ (s) + 2HNO₃ (aq)
From the balanced equation above,
2 moles of AgNO₃ reacted with 1 mole of H₂SO₄ to produce 1 mole of Ag₂SO₄ and 2 moles of HNO₃.
Next, we shall determine the limiting reactant.
This can obtained as follow:
From the balanced equation above,
2 moles of AgNO₃ reacted with 1 mole of H₂SO₄.
Therefore, 0.2 moles of AgNO₃ will react with = (0.2 x 1)/2 = 0.1 mole of H₂SO₄.
From the calculations made above, only 0.1 mole out of 0.155 mole of H₂SO₄ given is needed to react completely with 0.2 mole of AgNO₃. Therefore, AgNO₃ is the limiting reactant.
Next,, we shall determine the number of mole of Ag₂SO₄ produced from the reaction.
In this case we shall use the limiting reactant because it will give the maximum yield of Ag₂SO₄ as all of it is consumed in the reaction.
The limiting reactant is AgNO₃ and the number of mole of Ag₂SO₄ produced can be obtained as follow:
From the balanced equation above,
2 moles of AgNO₃ reacted to produce 1 mole of Ag₂SO₄.
Therefore, 0.2 moles of AgNO₃ will react to produce = (0.2 x 1)/2 = 0.1 mole of Ag₂SO₄.
Therefore, 0.1 mole of Ag₂SO₄ is produced from the reaction.
Finally, we shall convert 0.1 mole of Ag₂SO₄ to grams.
This can be obtained as follow:
Molar mass of Ag₂SO₄ = (2x108) + 32 + (16x4) = 312 g/mol
Mole of Ag₂SO₄ = 0.1
Mass of Ag₂SO₄ =?
Mole = mass /Molar mass
0.1 = Mass of Ag₂SO₄ /312
Cross multiply
Mass of Ag₂SO₄ = 0.1 x 312
Mass of Ag₂SO₄ = 31.2 g
Therefore, 31.2 g of Ag₂SO₄ were obtained from the reaction.
