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Marysya12 [62]
3 years ago
5

A mixture of 5L of H2 and 3L O2 reacts to form H2O (g) at constant T and P .Find the

Chemistry
1 answer:
slamgirl [31]3 years ago
4 0

The  volume of H₂O = 5 L

<h3>Further explanation</h3>

Given

5L of H₂ and 3L O₂

Reaction

2H₂ (g) + O₂(g)  ⇒2H₂O(g)

Required

The  volume of H₂O

Solution

Avogadro's hypothesis:  

<em>In the same T,P and V, the gas contains the same number of molecules  </em>

So the ratio of gas volume will be equal to the ratio of gas moles  

mol H₂ = 5, mol O₂ = 3

  • Find limiting reactants

From equation, mol ratio H₂ : O₂ = 2 : 1, so :

\tt H_2\div O_2=\dfrac{mol~H_2}{coefficient}\div \dfrac{mol~O_2}{coefficient}\\\\=\dfrac{5}{2}\div \dfrac{3}{1}=2.5\div 3\rightarrow H_2~limiting~reactant(smaller~ratio)

  • Find volume H₂O

mol H₂O based on mol H₂, and from equation mol ratio H₂ : H₂O=2 : 2, so mol H₂O = 5 mol and the volume also 5 L

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In a chemical reaction, exactly 2 mol of substance A react to produce exactly 3 mol of substance B. 2 A ⟶ 3 B How many molecules
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Answer:

There will be produced 1.71 moles of B which contain 1.03×10²⁴ molecules

Explanation:

The example reaction is:

2A → 3B

2 moles of A produce 3 moles of B

If we have the mass of A, we convert it to moles and then, we make the rule of three: 29.2 g / 25.6g/mol = 1.14 moles

Therefore 2 moles of A produce 3 moles of B

1.14 moles of A will produce (1.14 . 3) / 2 = 1.71 moles of B are produced

Now we can determine, the number of molecules

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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
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Answer:

31.2 g of Ag₂SO₄

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

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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:

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Mole of Ag₂SO₄ = 0.1

Mass of Ag₂SO₄ =?

Mole = mass /Molar mass

0.1 = Mass of Ag₂SO₄ /312

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Mass of Ag₂SO₄ = 0.1 x 312

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