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 water is produced by the reaction of 4.5 g of ethanol?
be sure your answer has the correct number of significant digits.

Explanation:

Step1: Write the balanced chemical equation

The reaction between ethanol (\(CH_3CH_2OH\)) and oxygen (\(O_2\)) to form water (\(H_2O\)) and acetic acid (\(CH_3COOH\)) is:

$$CH_3CH_2OH + O_2 ightarrow CH_3COOH + H_2O$$

Balancing the equation:

• Carbon: 2 on left, 2 on right (balanced).
• Hydrogen: 6 on left (\(CH_3CH_2OH\) has \(3 + 2 + 1 = 6\) H), 4 (from \(CH_3COOH\)) + 2 (from \(H_2O\)) = 6 on right (balanced).
• Oxygen: 1 (from \(CH_3CH_2OH\)) + 2 (from \(O_2\)) = 3 on left; 2 (from \(CH_3COOH\)) + 1 (from \(H_2O\)) = 3 on right. So balanced equation:

$$CH_3CH_2OH + O_2 ightarrow CH_3COOH + H_2O$$

Step2: Calculate moles of ethanol

Molar mass of \(CH_3CH_2OH\) (ethanol):

• C: \(2 \times 12.01 = 24.02\)
• H: \(6 \times 1.008 = 6.048\)
• O: \(1 \times 16.00 = 16.00\)

Total: \(24.02 + 6.048 + 16.00 = 46.068\) g/mol.

Moles of ethanol = \(\frac{\text{mass}}{\text{molar mass}} = \frac{4.5\ \text{g}}{46.068\ \text{g/mol}} \approx 0.09768\ \text{mol}\).

Step3: Relate moles of ethanol to water (mole ratio)

From the balanced equation, 1 mole of \(CH_3CH_2OH\) produces 1 mole of \(H_2O\). So mole ratio \(CH_3CH_2OH : H_2O = 1:1\).

Thus, moles of \(H_2O\) = moles of \(CH_3CH_2OH = 0.09768\ \text{mol}\).

Step4: Calculate mass of water

Molar mass of \(H_2O\): \(2 \times 1.008 + 16.00 = 18.016\) g/mol.

Mass of \(H_2O\) = moles \(\times\) molar mass = \(0.09768\ \text{mol} \times 18.016\ \text{g/mol} \approx 1.76\ \text{g}\).

Step5: Check significant digits

Mass of ethanol (4.5 g) has 2 significant digits. So the answer should have 2 significant digits.

\(1.76\ \text{g} \approx 1.8\ \text{g}\) (rounded to 2 significant digits).

Answer:

\(1.8\) g