Question

How many grams of O2(g) are needed to completely burn 92.2 g of C3H8(g)?

Answer 1

Answer on Question #59223 - Chemistry - Other

Task:

How many grams of $\mathrm{O}_2(\mathrm{g})$ are needed to completely burn $92.2\mathrm{g}$ of $\mathrm{C}_3\mathrm{H}_8(\mathrm{g})$?

Solution:

1) We find the molar mass of $\mathrm{C}_3\mathrm{H}_8$:

$$M \left(C _ {3} H _ {8}\right) = 3 \times 12 + 8 \times 1 = 36 + 8 = 44 (g / mol).$$

2) We find the amount of $\mathrm{C}_3\mathrm{H}_8$:

$$n \left(C _ {3} H _ {8}\right) = \frac {m \left(C _ {3} H _ {8}\right)}{M \left(C _ {3} H _ {8}\right)}.$$

Then,

$$n \left(C _ {3} H _ {8}\right) = \frac {92.2 \, \mathrm{g}}{44 \, \mathrm{g/mol}} = 2.095 \, \mathrm{mol}.$$

3) The reaction of burning of $\mathrm{C}_3\mathrm{H}_8$ is

$$\mathrm{C}_3\mathrm{H}_8 + 5\mathrm{O}_2 = 3\mathrm{CO}_2 + 4\mathrm{H}_2\mathrm{O}.$$

4) According to reaction the amount of oxygen is

$$n \left(O _ {2}\right) = 5 \times n \left(C _ {3} H _ {8}\right) = 5 \times 2.095 = 10.475 \, (\mathrm{mol}).$$

Then, the mass of oxygen is

$$m \left(O _ {2}\right) = n \left(O _ {2}\right) \times M \left(O _ {2}\right) = 10.475 \times 32 = 335.2 \, (\mathrm{g}).$$

Answer:

$$m \left(O _ {2}\right) = 335.2 \, \mathrm{g}$$

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