Question

when 11 moles of \\(\text{o}_2\\) reacts with 1.1 mole of \\(\text{c}_{10}\text{h}_8\\), what is the limiting reactant?

\\\text{c}_{10}\text{h}_8 + 12\text{o}_2 \
ightarrow 10\text{co}_2 + 4\text{h}_2\text{o}\\

• \\(\text{c}_{10}\text{h}_8\\)
• \\(\text{o}_2\\)
• \\(\text{co}_2\\)
• \\(\text{h}_2\text{o}\\)

Explanation:

Calculate the required moles of oxygen

$$ n(\text{O}_2)_{\text{required}} = 1.1\text{ mol C}_{10}\text{H}_8 \times \frac{12\text{ mol O}_2}{1\text{ mol C}_{10}\text{H}_8} = 13.2\text{ mol O}_2 $$

Compare available oxygen to required oxygen

$$
LATEXBLOCK0
$$

Identify the limiting reactant

$$ \text{Limiting Reactant} = \text{O}_2 $$

Answer:

• (A) \(\text{C}_{10}\text{H}_8\)
• (B) \(\text{O}_2\) (Correct answer)
• (C) \(\text{CO}_2\)
• (D) \(\text{H}_2\text{O}\)