Answer:
The limiting reactant is propene, [tex]C_3H_6[/tex].
Explanation:
[tex]2C_3H_6 + 9O_2 \rightarrow 6CO_2 + 6H_2O[/tex]
Moles of nitrogen propene = 2 mol
Moles of oxygen = 10 mol
According to reaction, 2 moles of propene reacts with 9 moles of oxygen gas, then 2 moles of propene will react with:
[tex]=\frac{9}{2}\times 2mol=9\text{mol of oxygen gas}[/tex]
According to the question, we have 10 moles of oxygen gas, which is more than 9 moles of oxygen gas. This indicates that propene is present in a limiting amount hence, it is a limiting reactant.
The limiting reactant is propene, hence the correct answer is the [tex]C_3H_6[/tex].
