Question

If you start with 11.8 g of C3H8 and 5.44 g of O2 which one is the limiting reagent?

Answer 1

Answer on the question #66415, Chemistry / Other

Question:

If you start with 11.8 g of C3H8 and 5.44 g of O2 which one is the limiting reagent?

Solution:

The reaction equation is:

$$C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$$

Thus, the relation of number of the moles of propane and oxygen is 1 to 5:

$$n(C_3H_8) = \frac{n(O_2)}{5}$$

Let's calculate the number of the moles of propane and oxygen from their masses:

$$n(C_3H_8) = \frac{m(C_3H_8)}{M(C_3H_8)} = \frac{11.8(g)}{44.0956(g \, \text{mol}^{-1})} = 0.2676 \, (\text{mol})$$ $$n(O_2) = \frac{m(O_2)}{M(O_2)} = \frac{5.44(g)}{31.9988(g \, \text{mol}^{-1})} = 0.1700 \, (\text{mol})$$

When we multiply the number of the moles of propane by 5, we get the number of the moles of oxygen required for complete combustion: $0.2676 \times 5 = 1.338$ (mol). The number of the moles of oxygen given is less:

$$0.1700 < 1.338$$

Thus, oxygen is the limiting reagent.

Answer:

Oxygen is the limiting reagent, as it's number of the moles is inferior to the number of the moles required for combustion of 11.8 g of propane.

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