Question

stoichiometry practice
use the following equation to answer # 1 - 9. \\( 2 \ce{c8h18} + 25 \ce{o2} \longrightarrow 16 \ce{co2} + 18 \ce{h2o} \\)

1. what kind of a reaction is this?
2. how many moles of \\( \ce{c8h18} \\) are required to produce 25 moles of \\( \ce{h2o} \\)?

1. how many grams of \\( \ce{co2} \\) are produced when 4 moles of \\( \ce{c8h18} \\) are reacted with excess oxygen?

1. how many moles of oxygen are required to produce 200g \\( \ce{h2o} \\)?

1. how many grams of \\( \ce{o2} \\) are needed to produce 250g of \\( \ce{co2} \\)?

1. how many grams of \\( \ce{c8h18} \\) are needed to form 4l of \\( \ce{co2} \\) gas?

Explanation:

Question 1

The reaction is $2 \ce{C8H18} + 25 \ce{O2}
ightarrow 16 \ce{CO2} + 18 \ce{H2O}$. A combustion reaction involves a hydrocarbon (here, $\ce{C8H18}$, an alkane) reacting with oxygen to produce carbon dioxide and water, which matches this reaction.

Step1: Identify mole ratio

From the balanced equation, $2$ moles of $\ce{C8H18}$ produce $18$ moles of $\ce{H2O}$. So the mole ratio of $\ce{C8H18}$ to $\ce{H2O}$ is $\frac{2}{18}$.

Step2: Calculate moles of $\ce{C8H18}$

Let $x$ be moles of $\ce{C8H18}$. We have $\frac{2}{18}=\frac{x}{25}$. Solving for $x$: $x = \frac{2\times25}{18}=\frac{50}{18}\approx2.78$ moles.

Step1: Mole ratio of $\ce{C8H18}$ to $\ce{CO2}$

From the equation, $2$ moles of $\ce{C8H18}$ produce $16$ moles of $\ce{CO2}$. So for $4$ moles of $\ce{C8H18}$, moles of $\ce{CO2}$: $\frac{16}{2}\times4 = 32$ moles.

Step2: Molar mass of $\ce{CO2}$

Molar mass of $\ce{CO2}$: $12 + 2\times16 = 44$ g/mol.

Step3: Mass of $\ce{CO2}$

Mass = moles × molar mass = $32\times44 = 1408$ g.

Answer:

Combustion reaction (specifically, hydrocarbon combustion)

Question 2