How many moles of water are produced when 5 moles of hydrogen gas react with 2 moles of oxygen gas?

Balanced equation: 2H2 + O2 → 2H2O (5 points)


a. 2 moles of water
b. 4 moles of water
c. 5 moles of water
d. 7 moles of water

Answers

Answer 1

Answer:

It is b

Explanation: I did it on the test and got it right

Answer 2

The moles of water are produced when 5 moles of hydrogen gas react with 2 moles of oxygen gas - b. 4 moles of water

The stoichiometric coefficients can be considered in a balanced equation as moles of reactants needed or moles of products formed in the reaction.  

In the given case, the balanced chemical equation for this synthesis reaction looks like this

2H2 + O2 → 2H2O  

from the balanced equation, it is clear that the 2  moles of hydrogen gas and 1  mole of oxygen gas are required to produce  2  moles of water.  so it means that the reaction produces twice as many moles of water as you have moles of oxygen gas that take part in the reaction and the same moles as hydrogen atoms present.Here it is given that there are only 2 moles of oxygen which means it is in deficiency however, it will produce 4 moles of water with four moles of hydrogen.

Thus, the moles of water are produced when 5 moles of hydrogen gas react with 2 moles of oxygen gas - b. 4 moles of water

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Nitrous oxide

Names

Molar mass 44.013 g/mol

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How would adding the catalyst nitrogen monoxide (NO) affect this reaction?

2SO2(g) + O2(g) → 2SO3(g)

A) NO increases the rate at which SO3 molecules are formed.

B) NO reacts with SO3 to produce more SO2 molecules.

C) NO decreases collisions between the SO2 and O2 molecules.

D) NO increases the concentration of the SO2 and O2 molecules.

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Answers

Answer: A) NO increases the rate at which [tex]SO_3[/tex] molecules are formed.

Explanation:

A catalyst is a substance which increases the rate of a reaction by taking the reaction through a different path which involves lower activation energy and thus more reactant molecules can cross the energy barrier by undergoing collisions and convert to products.

[tex]2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)[/tex]

Thus NO will increase the rate of reaction by lowering the activation energy and thus the colllisions among [tex]SO_2[/tex] and [tex]O_2[/tex] molecules will incraese which in turn will lead to formatioon of more [tex]SO_3[/tex] molecules.

Answer:

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Explanation:

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Explanation:

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Answer:

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Explanation:

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True

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Answers

1.563 g

Further explanation

Given

A half life of 2.5 days ⇒ t1/2 = 2.5 days

initial sample ⇒ 25 g

duration of decay ⇒ 10 days

Required

sample left after 10 days

Analysis

Use decay Formula :

[tex]\tt Nt=No.\dfrac{1}{2}^{T/t1/2}[/tex]

Solution

[tex]\tt Nt=25.\dfrac{1}{2}^{10/2.5}\\\\Nt=25.\dfrac{1}{2}^4\\\\Nt=\boxed{\bold{1.563~g}}[/tex]

Paraphrase

Substance left after 10 days = 1.563 g

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