Question

How many grams of oxygen are required to burn 4.4 gram of C3 h8?

Answer 1

Answer on Question #80184 – Chemistry – Other

**Task:**

How many grams of oxygen are required to burn 4.4 gram of $C_3H_8$?

**Solution:**

Balanced equation of combustion of propane:

$$C_3H_8 + 5O_2 = 3CO_2 + 4H_2O$$

By the reaction equation:

$$n(C_3H_8) = \frac{n(O_2)}{5};$$ $$\frac{m(C_3H_8)}{M(C_3H_8)} = \frac{m(O_2)}{5 * M(O_2)};$$ $$m(O_2) = \frac{5 * M(O_2) * m(C_3H_8)}{M(C_3H_8)} = \frac{5 * (2 * 16) * 4.4}{(12 * 3 + 1 * 8)} = \frac{704}{44} = 16g$$

Answer: 16 grams of oxygen.

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