Question

for the following reaction, 0.339 moles of butane (c₄h₁₀) are mixed with 0.296 moles of oxygen gas. butane (c₄h₁₀)(g)+oxygen(g)→carbon dioxide(g)+water(g) what is the formula for the limiting reagent? limiting reagent: what is the maximum amount of carbon dioxide that can be produced? amount = moles

Explanation:

Step1: Balance the chemical equation

$2C_4H_{10}(g)+13O_2(g)
ightarrow8CO_2(g) + 10H_2O(g)$

Step2: Calculate the moles of $O_2$ required to react with given $C_4H_{10}$

From the balanced equation, the mole - ratio of $C_4H_{10}$ to $O_2$ is $2:13$. Given $n_{C_4H_{10}} = 0.339$ moles. The moles of $O_2$ required, $n_{O_2 - required}=\frac{13}{2}\times0.339= 2.2035$ moles. But we have only $0.296$ moles of $O_2$.

Step3: Determine the limiting reagent

Since the amount of $O_2$ available is less than the amount required to react with all of the $C_4H_{10}$, $O_2$ is the limiting reagent.

Step4: Calculate the moles of $CO_2$ produced

The mole - ratio of $O_2$ to $CO_2$ is $13:8$. Let the moles of $CO_2$ produced be $x$. Using the proportion $\frac{x}{0.296}=\frac{8}{13}$, we get $x=\frac{8\times0.296}{13}\approx0.182$ moles.

Answer:

Limiting reagent: $O_2$
Amount = $0.182$ moles