Question

Liquid hexane
will react with gaseous oxygen
to produce gaseous carbon dioxide
and gaseous water
. Suppose 13.8 g of hexane is mixed with 74. g of oxygen. Calculate the minimum mass of hexane that could be left over by the chemical reaction. Round your answer to
significant digits.

Answer 1

Answer on Question #80337, Chemistry/ Organic Chemistry

Liquid hexane

will react with gaseous oxygen

to produce gaseous carbon dioxide

and gaseous water

. Suppose 13.8 g of hexane is mixed with 74. g of oxygen. Calculate the minimum mass of hexane that could be left over by the chemical reaction. Round your answer to significant digits.

Solution

$$2 C _ {6} H _ {1 4} (l) + 1 9 O _ {2} (g) \rightarrow 1 2 C O _ {2} (g) + 1 4 H _ {2} O (g)$$

1. Find amount of substance of hexane:

$$\begin{array}{l} n = m / M \\ M \left(C _ {6} H _ {1 4}\right) = 8 6 \mathrm {g} / \mathrm {m o l} \\ n \left(C _ {6} H _ {1 4}\right) = 1 3. 8 \mathrm {g} / 8 6 \mathrm {g} / \mathrm {m o l} = 0. 1 6 0 \mathrm {m o l} \\ \end{array}$$

2. Find amount of substance of oxygen

$$\begin{array}{l} n = m / M \\ M \left(O _ {2}\right) = 3 2 \mathrm {g} / \mathrm {m o l} \\ n \left(O _ {2}\right) = 7 4. \mathrm {g} / 3 2 \mathrm {g} / \mathrm {m o l} = 2. 3 1 \mathrm {m o l} \\ \end{array}$$

3. Find limiting reactant.

According to equation mole ratio of reactants is:

$$n \left(C _ {6} H _ {1 4}\right): n \left(O _ {2}\right) = 2: 1 9.$$

If we have 0.160 mol of hexane then amount of substance of oxygen should be: $n(O_2) = 0.160 \times 19/2 = 1.52$ mol. And we have $n(O_2) = 2.31$ mol. We have excess of oxygen. Limiting reactant is hexane. This means, that all hexane (13.8 g) will be consumed in this reaction. No hexane could be left over.

Answer: all hexane (13.8 g) will be consumed in this reaction, no hexane could be left over.