Respuesta :
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.
Taking into account the definition of reaction stoichiometry and limiting reagent, the mass of Ag₂SO₄ that could be formed is 31.18 grams.
First of all, the balanced reaction is:
2 AgNO₃ + H₂SO₄ → Ag₂SO₄ + 2 HNO₃
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- AgNO₃: 2 moles
- H₂SO₄: 1 mole
- Ag₂SO₄: 1 mole
- HNO₃: 2 moles
The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.
To determine the limiting reagent, you can use a simple rule of three as follows: if by stoichiometry 1 mole of H₂SO₄ reacts with 2 moles of AgNO₃, 0.155 moles of H₂SO₄ react with how many moles of AgNO₃?
[tex]amount of moles of AgNO_{3} =\frac{0.155 moles of H_{2}SO_{4} x 2 moles of AgNO_{3} }{1 mole of H_{2}SO_{4} }[/tex]
moles of AgNO₃= 0.31 moles
But 0.31 moles of AgNO₃ are not available, 0.200 moles are available. Since you have less moles than you need to react with 0.155 moles of H₂SO₄, AgNO₃ will be the limiting reagent.
Then, it is possible to determine the amount of moles of Ag₂SO₄ produced by another rule of three, using the limiting reagent: if by stoichiometry 2 moles of AgNO₃ produce 1 mole of Ag₂SO₄, 0.200 moles of AgNO₃ how many moles of Ag₂SO₄ will be formed?
[tex]amount of moles of Ag_{2} SO_{4} =\frac{1 mole of Ag_{2} SO_{4} x 0.200 moles of AgNO_{3} }{2 moles of AgNO_{3} }[/tex]
amount of moles of Ag₂SO₄ =0.100 moles
Finally, with 311.8 g/mole being the molar mass of Ag₂SO₄, then the mass produced of the compound can be calculated as:
[tex]0.100 molesx311.8 \frac{g}{mole} = 31.18 grams[/tex]
In summary, the mass of Ag₂SO₄ that could be formed is 31.18 grams.
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