Question

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 oxygen gas is consumed by the reaction of 7.6g of ethanol? round your answer to 2 significant digits.

Answer 1

The mass of oxygen required to react with 7.6 g of ethanol is 5.3 g.

Further Explanation

This problem is a stoichiometric problem. To solve problems like this we always begin with the balanced equation. Then, we use the stoichiometric ratios provided by the coefficients of the reactants and products.

1. Write the balanced chemical equation.

CH₃CH₂OH + O₂ → H₂O + CH₃COOH

2. Convert the mass of oxygen to moles.

$7.6 \ g \ CH_3CH_2OH \times \frac{1 \ mol CH_3CH_2OH}{46.07 \ g} = 0.165 \ mol \ CH_3CH_2OH$

3. Determine the equivalent moles of oxygen that reacts with the given quantity of ethanol. The stoichiometric ratio for ethanol and oxygen indicated in the balanced chemical equation is 1:1.

$moles \ O_2 = 0.165 \ mol \ CH_3CH_2OH \times \frac{1 \ mol O_2}{ 1 \ mol CH_3CH_2OH}\\\\moles \ O_2 = 0.165 \ mol \ O_2$

4. Convert the moles of O₂ to mass.

$mass \ of \ O_2 = 0.165 \ mol \ O_2 \times \frac{32 \ g}{1 \ mol \ O_2 }\\\\\boxed {mass \ of \ O_2 \ = 5.28 \ g}$

The answer required should only have 2 significant digits. Therefore, the final answer is:

$\boxed {\boxed {mass \ of \ O_2 \ = 5.3 \ g}}$

Learn More

1. Learn more about mole to particle conversion brainly.com/question/12972204
2. Learn more about stoichiometry and excess brainly.com/question/12979299

Keywords: stoichiometry, fermentation

Answer 2

1) Chemical reaction

CH3 CH2 OH + O2 ---> CH3COOH + H2O

It is balanced

2) Molar ratios

1 mol CH3 CH2OH : 1 mol O2 : 1 mol CH3COOH : 1 mol H2O

3) Conversion of 7.6 grams of ethanol to moles

Molar mass of CH3CH2OH = 2 * 12g/mol + 6 * 1 g/mol + 16 g/mol = 46 g/mol

moles = mass / molar mass = 7.6 g / 46g/mol = 0.165 mol

4) Use of proportions with the molar ratios:

1mol CH3CH2OH / 1mol O2 = 0.165 mol CH3CH2OH / x => x = 0.165 mol CH3CH2OH.

5)  Conversion of 0.165 mol O2 to grams

mass in grams = number of moles * molar mass = 0.165 mol * (2 * 16g/mol)

mass = 5.28 g ≈ 5.3 g

Answer: 5.3 g