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Nana76 [90]
3 years ago
11

4NH3(g)+5O2(g)⟶4NO(g)+6H2O(g) how many molecules of H2O can be formed from 6.12mol NH3?

Chemistry
1 answer:
stich3 [128]3 years ago
5 0

Answer:

The answer to your question is 5.529 x 10²⁴ molecules of H₂O

Explanation:

                     4NH₃  + 5O₂   ⇒   4NO   +   6H₂O

                      6.12 mol                               ? molecules

Rule of three (proportions)

                     4 mol of NH₃  ------------------- 6 mol of H₂O

                     6.12 mol of NH₃  ---------------  x mol of H₂O

                     x = (6.12 x 6) / 4

                     x = 36.72 / 4

                     x = 9.18 mol of H₂O

                           

                     1 mol of H₂O ------------------ 6.023 x 10²³ molecules

                     9.18 mol of H₂O -------------   x

                     x = (9.18 x 6.023 x 10²³) / 1

                     x = 5.529 x 10²⁴ molecules

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Question 2 of 50
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The thermal decomposition of calcium carbonate will produce 14 g of calcium oxide. The stoichiometric ratio of calcium carbonate to calcium oxide is 1:1, therefore the number of moles of calcium carbonate decomposed is equal to the number of moles of calcium oxide formed.

Further Explanation:

To solve this problem, follow the steps below:

  1. Write the balanced chemical equation for the given reaction.
  2. Convert the mass of calcium carbonate into moles.
  3. Determine the number of moles of calcium oxide formed by using the stoichiometric ratio for calcium oxide and calcium carbonate based on the coefficient of the chemical equation.
  4. Convert the number of moles of calcium oxide into mass.

Solving the given problem using the steps above:

STEP 1: The balanced chemical equation for the given reaction is:

CaCO_{3} \rightarrow \ CaO \ + \ CO_{2}

STEP 2: Convert the mass of calcium carbonate into moles using the molar mass of calcium carbonate.

mol \ CaCO_{3} \ = 25 \ g \ CaCO_{3} \ (\frac{1 \ mol \ CaCO_{3}}{100.0869 \ g \ CaCO_{3}})\\ \\\boxed {mol \ CaCO_{3} \ = 0.2498 \ mol}

STEP 3: Use the stoichiometric ratio to determine the number of moles of CaO formed.

For every mole of calcium carbonate decomposed, one more of a calcium oxide is formed. Therefore,

mol \ CaO \ = 0.2498 \ mol

STEP 4: Convert the moles of CaO into mass of CaO using its molar mass.

mass \ CaO \ = 0.2498 \ mol \ CaO \ (\frac{56.0774 \ g \ CaO}{1 \ mol \ CaO})\\ \\mass \ CaO \ = 14.008 \ g

Since there are only 2 significant figures in the given, the final answer must have the same number of significant figures.

Therefore,

\boxed {mass \ CaO \ = 14 \ g}

Learn More

  1. Learn more about stoichiometry brainly.com/question/12979299
  2. Learn more about mole conversion brainly.com/question/12972204
  3. Learn more about limiting reactants brainly.com/question/12979491

Keywords: thermal decomposition, stoichiometry

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Answer:

Explanation has been given below.

Explanation:

  • Chloroform has three polar C-Cl bonds. Methylene chloride has two polar C-Cl bonds. So it is expected that chloroform should be more polar and posses higher dipole moment than methylene chloride.
  • Two factors are liable for the opposite trend observed in dipole moments of methylene chloride and chloroform.
  • First one is the number of hyperconjugative hydrogen atoms present in a molecule. Hyperconjugation occurs with vacant d-orbital of Cl atom. Hyperconjugation amplifies charge separation in a molecule resulting higher dipole moment.
  • Methylene chloride has two hyperconjugative hydrogen atoms and chloroform has one hyperconjugative hydrogen atom.Therefore methylene chloride should have higher charge separation as compared to chloroform.
  • Second one is induction of opposite polarity in a C-Cl bond by another C-Cl bond in a molecule. Higher the opposite induction of polarity, lower the charge separation in a molecule and hence lower the dipole moment of a molecule.
  • Chloroform has three C-Cl bonds and methylene chloride has two C-Cl bonds. Therefore opposite induction is higher for chloroform resulting it's lower dipole moment.
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