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solniwko [45]
2 years ago
13

Wine goes bad soon after opening because the ethanol (CH,CH,OH) in it reacts with oxygen gas (0.) from the air to form water (11

0) and acetic acid (CH,COOH), the main ingredient of vinegar. What mass of water is produced by the reaction of 4.59 g of ethanol Be sure your answer has the correct number of significant digits.
Chemistry
1 answer:
daser333 [38]2 years ago
5 0

<u>Answer:</u> The mass of water produced is 1.8 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

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

Given mass of ethanol = 4.59 g

Molar mass of ethanol = 46.07 g/mol

Putting values in equation 1, we get:

\text{Moles of ethanol}=\frac{4.59g}{46.07g/mol}=0.0996mol

The chemical equation for the reaction of ethanol with oxygen gas follows:

CH_3CH_2OH+O_2\rightarrow CH_3COOH+H_2O

By Stoichiometry of the reaction;

1 mole of ethanol produces 1 mole of water.

So, 0.0996 moles of ethanol will produce = \frac{1}{1}\times 0.0996=0.0996mol of water.

Now, calculating the mass of water from equation 1, we get:

Molar mass of water = 18 g/mol

Moles of water = 0.0996 moles

Putting values in equation 1, we get:

0.0996mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(0.0996mol\times 18g/mol)=1.8g

Hence, the mass of water produced is 1.8 grams

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The pressure of the gas = 40 atm

<h3>Further explanation</h3>

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\tt \rm p_1V_1=p_2.V_2\\\\\dfrac{p_1}{p_2}=\dfrac{V_2}{V_1}

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A 3.4 g sample of an unknown monoprotic organic acid composed of C,H, and O is burned in air to produce 8.58 grams of carbon dio
Pavlova-9 [17]

Answer:

C_7H_6O_2

Explanation:

Hello there!

In this case, we can divide the problem in three stages: (1) determine the empirical formula with the combustion analysis, (2) compute the molar mass of acid via the moles of the acid in the neutralization and (3) determine the molecular formula.

(1) In this case, since 8.58 g of carbon dioxide are released, we can first compute the moles of carbon in the compound:

n_C=8.58gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.195molC

And the moles of hydrogen due to the produced 1.50 grams of water:

n_H=1.50gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O}  =0.166molH

Next, to compute the mass and moles of oxygen, we need to use the initial 3.4 g of the acid:

m_O=3.4g-0.195molC*\frac{12.01gC}{1molC}-0.166molH*\frac{1.01gH}{1molH} =0.89gO\\\\n_O=0.89gO*\frac{1molO}{16.0gO}=0.0556molO

Thus, the subscripts in the empirical formula are:

C=\frac{0.195}{0.0556}=3.5 \\\\H=\frac{0.166}{0.0556}=3\\\\O=\frac{0.0556}{0.0556}=1\\\\C_7H_6O_2

As they cannot be fractions.

(2) In this case, since the acid is monoprotic, we can compute the moles by multiplying the concentration and volume of KOH:

n_{KOH}=0.279L*0.1mol/L\\\\n_{KOH}=0.0279mol

Which are equal to the moles of the acid:

n_{acid}=0.0279mol

And the molar mass:

MM_{acid}=\frac{3.4g}{0.0279mol} =121.86g/mol

(3) Finally, since the molar mass of the empirical formula is:

7*12.01 + 6*1.01 + 2*16.00 = 122.13 g/mol

Thus, since the ratio of molar masses is 122.86/122.13 = 1, we infer that the empirical formula equals the molecular one:

C_7H_6O_2

Best regards!

8 0
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