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Aleonysh [2.5K]
2 years ago
7

How many moles of O2 are required to react with 2.4 mol of H2 ?

Chemistry
1 answer:
Leokris [45]2 years ago
8 0

Answer:

1.2 moles

Explanation:

this is the balanced equation for the reaction of oxygen (O2) and hydrogen (H2), usually we don't write the 1 in front of O2

2H₂ + 10₂ → 2H₂O

the molar ratio of hydrogen to oxygen is 2 : 1

we are trying to react with 2.4 mol of H2 so the moles of O2 is half the number of moles of H2 = 2.4 ÷ 2 = 1.2 mol

another way to think of it:

2H₂ + 10₂

2 : 1

2.4 mol : x mol

to get from 2 to 2.4 multiply by 1.2, so do the same to the other side

1 × 1.2 = 1.2 mol

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Therefore, according to the question, the procedure that best demonstrates the law of conservation of mass after the chemistry student conducted different procedures is option D.

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You conduct an experiment that requires the creation of an ammonia solution. You do this by reacting 50.0 L of nitrogen gas with
ozzi

Answer : The molarity of the resulting ammonia solution is, 0.89 M

Explanation :

The balanced chemical reaction is:

N_2(g)+3H_2(g)\rightarrow 2NH_3(g)

First we have to calculate the moles of nitrogen gas.

As we know that at STP, 1 mole of gas occupies 22.4 L volume of gas.

As, 22.4 L volume of nitrogen gas present in 1 moles of nitrogen gas

So, 50.0 L volume of nitrogen gas present in \frac{50.0}{22.4}=2.23 moles of nitrogen gas

Thus, the moles of nitrogen gas is 2.23 moles.

Now we have to calculate the moles of ammonia gas.

From the reaction, we conclude that

As, 1 mole of N_2 react to give 2 mole of NH_3

So, 2.23 moles of N_2 react to give 2.23\times 2=4.46 moles of NH_3

Now we have to calculate the molarity of the resulting ammonia solution.

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

\text{Molarity}=\frac{\text{Moles of ammonia}}{\text{Volume of solution (in L)}}

Now put all the given values in this formula, we get:

\text{Molarity}=\frac{4.46mole}{5.0L}

\text{Molarity}=0.89M

Therefore, the molarity of the resulting ammonia solution is, 0.89 M

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3 years ago
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