Question

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 are 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

Answer 1

Answer:

• Last choice: 109.729 g

Explanation:

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

The combustion reaction of isopropyl alcohol 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 $H_2O$.

1. Convert the 91.5 g of isopropyl alcohol into number of moles

• 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

2. Calculate the the number of moles of water produced using the mole ratio:

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

3. Convert to mass of water

• 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.