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koban [17]
3 years ago
13

Shana solves stoichiometry problems using the equation for the synthesis of water. Which interpretation of the balanced equation

would cause Shana to make a mistake? Two moles of hydrogen react with one mole of oxygen to form two moles of water. Two grams of hydrogen react with one gram of oxygen to form two grams of water. Two molecules of hydrogen react with one molecule of oxygen to form two molecules of water. Two liters of hydrogen react with one liter of oxygen to form two liters of water.
Chemistry
2 answers:
vampirchik [111]3 years ago
8 0

Answer : The wrong statement is, Two grams of hydrogen react with one gram of oxygen to form two grams of water.

Explanation :

Synthesis reaction : It is a type of reaction in which the reactants present in elemental state that reacts to give a single product.

It is represented as,

X+Y\rightarrow XY

where, X and Y are the reactants and XY is the product.

The balanced synthesis reaction of water will be,

2H_2+O_2\rightarrow 2H_2O

By the stoichiometry we can say that, 2 moles of hydrogen (H_2) react with 1 mole of oxygen (O_2) to form 2 moles of water (H_2O) as a single product.

Or we can say that,

2 molecules of hydrogen react with 1 molecule of oxygen to form 2 molecules of water.

Or,

2 liters of hydrogen react with 1 liter of oxygen to form 2 liters of water.

These three statements are correct.

But the statement, 2 grams of hydrogen react with 1 gram of oxygen to form 2 grams of water are incorrect statement.

Hence, the wrong statement is, Two grams of hydrogen react with one gram of oxygen to form two grams of water.

skelet666 [1.2K]3 years ago
6 0

Answer: Two grams of hydrogen react with one gram of oxygen to form two grams of water.

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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
eduard

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%.

5 0
3 years ago
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