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Harrizon [31]
3 years ago
5

An experiment requires that enough C3H8O be used to yield of oxygen .

Chemistry
1 answer:
blsea [12.9K]3 years ago
8 0

Answer: So if you had 570 cm of ribbon, then 570%2F8.5=67.05 which means that about 67 students can do the experiment (round down to the nearest whole number).

Explanation: If you had 8.5 cm of ribbon, then only 8.5%2F8.5=1 student can do the experiment. If you had 17 cm of ribbon, then 17%2F8.5=2 students can do the experiment.

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During a laboratory experiment, a 2.36-gram sample of NaHCO3 was thermally decomposed. In this experiment, carbon dioxide and wa
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Answer:

  • 90.7 %

Explanation:

<u>1) Chemical equation (given)</u>

  • 2NaHCO₃ → Na₂CO₃ + H₂CO₃

<u>2) Theoretical yield</u>

<u>a) Convert mass of NaHCO₃ to moles:</u>

  • n = mass in grams / molar mass
  • molar mass = 84.007 g/mol
  • n = 2.36 g / 84.007 g/mol = 0.02809 mol

<u>b) Mole ratio:</u>

  • 2 mol NaHCO₃ : 1 mol H₂CO₃

<u>c) Proportionality:</u>

  • 2 mol NaHCO₃ / mol H₂CO₃ = 0.02809 mol NaHCO₃ / x

       ⇒ x = 0.2809 / 2 mol H₂CO₃ = 0.01405 mol H₂CO₃

<u>3) Actual yield</u>

<u>a) Mass balance</u>: 2.36 g - 1.57 g = 0.79 g

<u>b) Convert 0.79 g of carbonic acid to number of moles</u>:

  • n = mass in grams / molar mass

  • molar mass = 62.03 g/mol

  • n = 0.79 g / 62.03 g/mol = 0.01274 mol

<u>4) Percentage yield, y (%)</u>

  • y (%) = actual yield / theoretical yield × 100

  • y (%) = 0.1274 mol / 0.1405 mol × 100 = 90.68%

The answer must show 3 significant figures, so y(%) = 90.7%.

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