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levacccp [35]
3 years ago
9

When 91.5 g of isopropyl alcohol which has an empirical formula of C3H8O is burned in excess oxygen gas, how many grams of H2O a

re formed? MWC = 12.011 g/mol, MWH = 1.00794 g/mol, and MWO = 15.9994 g/mol.
1. 47.9229
2. 84.1255
3. 52.2617
4. 49.8948
5. 86.9152
6. 119.705
7. 88.0758
8. 76.2076
9. 62.9663
10. 109.729
Chemistry
1 answer:
Snowcat [4.5K]3 years ago
8 0

Answer:

  • <u><em>Last choice: 109.729 g</em></u>

Explanation:

This is a combustion reaction. The complete combustion of a hydrocarbon produces carbon dioxide and water.

The combustion reaction of<em> isopropyl alcohol</em> is represented by the balanced hemical equation:

            2C_3H_8O+9O_2\rightarrow 6CO_2+8H_2O

Thus, 2 moles of C_3H_8O produce 8 moles of H_2O.

<u>1. Convert the 91.5 g of isopropyl alcohol into number of moles</u>

  • Number of moles = molar mass / fomula mass

  • Formula mass

 3\times 12.011g/mol+8\times 1.00794g/mol+1\times 15.9994g/mol=60.09592g/mol

  • Number of moles = 91.5g / (60.95592g/mol) = 1.52256692 mol

<u>2. Calculate the the number of moles of water produced using the mole ratio:</u>

         \dfrac{8molH_2O}{2molC_3H_8O}\times 1.52256692molC_3H_8O=6.0902637molH_2O

<u />

<u>3. Convert to mass of water</u>

  • Mass = number of moles × molar mass
  • Molar mass = 2 × 1.00794 g/mol + 15.9994g/mol = 18.01528‬g/mol
  • Mass = 6.0902637 mol × 18.01528 g/mol = 109.717806g ≈ 109.72 g

So far, I have kept a large number of digits because these are intermediate calculations.

You should report the result with 3 significant figures, but all the answer choices contain more than that.

The last choice, 109.729, is the closest to the correct value.

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