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Arisa [49]
3 years ago
6

Wine goes bad soon after opening because the ethanol CH3CH2OH in it reacts with oxygen gas O2 from the air to form water H2O and

acetic acid CH3COOH, the main ingredient of vinegar. What mass of ethanol is consumed by the reaction of 1.9g of oxygen gas?
Chemistry
1 answer:
lyudmila [28]3 years ago
3 0

Answer: 2.73g of CH3CH2OH Will be consumed

Explanation:

CH3CH2OH + O2 —> CH3COOH + H2O

MM of CH3CH2OH = 12 + 3 +12 + 2 16 +1 = 46g/mol

MM of O2 = 16 x2 = 32

Mass conc. Of O2 = 1.9g

From the equation,

32g of O2 consumed 46g of CH3CH2OH .

Therefore, 1.9g of O2 will consume Xg of CH3CH2OH i.e

Xg of CH3CH2OH = (1.9 x 46)/32 = 2.73g

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Now we need to find the amount of NF3 that can be formed by the complete reactions of each of the reactants. If all of the N2 wa
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The question is incomplete, the complete question is:

Nitrogen and fluorine react to form nitrogen fluoride according to the chemical equation:

N_2(g)+3F_2(g)\rightarrow 2NF_3(g)

A sample contains 19.3 g of N_2 is reacted with 19.3 g of F_2. Now we need to find the amount of NF_3 that can be formed by the complete reactions of each of the reactants.

If all of the N_2 was used up in the reaction, how many moles of NF_3 would be produced?

<u>Answer:</u> 1.378 moles of NF_3 are produced in the reaction.

<u>Explanation:</u>

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}       ......(1)

Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.

Excess reagent is defined as the reagent which is left behind after the completion of the reaction.

In the given chemical reaction, N_2 is considered as a limiting reagent because it limits the formation of the product and it was completely consumed in the reaction.

We are given:

Mass of N_2 = 19.3 g

Molar mass of N_2 = 28.02 g/mol

Putting values in equation 1:

\text{Moles of }N_2=\frac{19.3g}{28.02g/mol}=0.689mol

For the given chemical reaction:

N_2(g)+3F_2(g)\rightarrow 2NF_3(g)

By the stoichiometry of the reaction:

1 mole of N_2 produces 2 moles of NF_3

So, 0.689 moles of N_2 will produce = \frac{2}{1}\times 0.689=1.378mol of NF_3

Hence, 1.378 moles of NF_3 are produced in the reaction.

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