Question

acetylene (c₂h₂) gas and oxygen (o₂) gas react to form carbon dioxide (co₂) gas and water (h₂o) vapor. suppose you have 13.0 mol of c₂h₂ and 1.0 mol of o₂ in a reactor. suppose as much as possible of the c₂h₂ reacts. how much will be left? round your answer to the nearest 0.1 mol.

Explanation:

Step1: Write the balanced chemical equation

$2C_2H_2 + 5O_2
ightarrow4CO_2 + 2H_2O$

Step2: Determine the mole - ratio

The mole - ratio of $C_2H_2$ to $O_2$ is $2:5$.

Step3: Calculate the amount of $C_2H_2$ that reacts with $1.0$ mol of $O_2$

Let $x$ be the amount of $C_2H_2$ that reacts. Using the mole - ratio $\frac{x}{1.0\ mol\ O_2}=\frac{2}{5}$, so $x=\frac{2\times1.0\ mol\ O_2}{5}=0.4\ mol\ C_2H_2$.

Step4: Calculate the amount of $C_2H_2$ left

Initial amount of $C_2H_2$ is $13.0\ mol$. Amount of $C_2H_2$ that reacts is $0.4\ mol$. So the amount of $C_2H_2$ left is $n = 13.0\ mol-0.4\ mol = 12.6\ mol$.

Answer:

$12.6$