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sveticcg [70]
3 years ago
10

If 0.200 moles of AgNO₃ react with 0.155 moles of H₂SO₄ according to this UNBALANCED equation below, what is the mass in grams o

f Ag₂SO₄ that could be formed? AgNO₃(aq) + H₂SO₄ (aq) → Ag₂SO₄ (s) + HNO₃ (aq)
Chemistry
1 answer:
morpeh [17]3 years ago
5 0

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.

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3 years ago
3 Cu + 8HNO3 g 3 Cu(NO3)2 + 2 NO + 4 H2O
liubo4ka [24]

24.25 moles of NO can be produced using 97 moles of HNO3.

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3 Cu + 8HNO3 g → 3 Cu(NO3)2 + 2 NO + 4 H2O

The number of moles consumed can be calculated using comparing with coefficients in the balanced reaction .

So , from above eq we get that 8 moles of HNO3 are consumed to make 2 moles of NO.

⇒  8 HNO3⇔2 NO

⇒ 1 HNO3⇔ 1/4 NO

This means that for each mole of HNO3 produces 1/4 moles of NO.

So , for 97 moles of HNO3 , \frac{1}{4}  *97 moles of NO can be made,

So, total moles of NO made are 24.25 moles.

Lean more about balanced reactions here brainly.com/question/26694427

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