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Methanol (CH3OH) burns in air according to the equation If

Chapter 3, Problem 13PE

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QUESTION:

Methanol \(\left(C H_{3} O H\right)\) burns in air according to the equation

\(2 \mathrm{CH}_{3} \mathrm{OH}+3 \mathrm{O}_{2} \rightarrow 2 \mathrm{CO}_{2}+4 \mathrm{H}_{2} \mathrm{O}\)

If 209 g of methanol are used up in a combustion process, what is the mass of \(\mathrm{H}_{2} \mathrm{O}\) produced?

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QUESTION:

Methanol \(\left(C H_{3} O H\right)\) burns in air according to the equation

\(2 \mathrm{CH}_{3} \mathrm{OH}+3 \mathrm{O}_{2} \rightarrow 2 \mathrm{CO}_{2}+4 \mathrm{H}_{2} \mathrm{O}\)

If 209 g of methanol are used up in a combustion process, what is the mass of \(\mathrm{H}_{2} \mathrm{O}\) produced?

ANSWER:

Step 1 of 2

The goal of the problem is to find the mass of  produced in the given reaction.

Given reaction:

Given:

Mass of methanol  = 209 g

Here, methanol is the limiting reagent. Let’s calculate the moles of methanol and find the corresponding quantity of water that is formed.

First, let’s determine the number of moles in methanol:

Molecular mass of methanol = 32.04 g/mol.

Number of moles =

                  n =

     = 6.52 moles  

We see in the balanced equation that ratio of methanol to water is  2: 4. Therefore moles of  formed is:

= 6.52 moles of   

= 13.04 moles of  

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