Question

Liquid octane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 10.3 g of octane is mixed with 69. g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Round your answer to significant digits.

Answer 1

Answer:

$\large \boxed{\text{31.8 g CO}_{2}}$

Explanation:

We are given the masses of two reactants and asked to determine the mass of the product.

This looks like a limiting reactant problem.

1. Assemble the information

We will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.  

MM:         114.23     32.00      44.01

               2C₈H₁₈ + 25O₂ ⟶ 16CO₂ + 18H₂O

Mass/g:     10.3        69.  

2. Calculate the moles of each reactant

$\text{Moles of C _{8} H _{18} } = \text{10.3 g C _{8} H _{18} } \times \dfrac{\text{1 mol C _{8} H _{18} }}{\text{114.23 mol C _{8} H _{18} }} = \text{0.090 17 mol C _{8} H _{18} }\\\\\text{Moles of O _{2} } = \text{69. g O}_{2} \times \dfrac{\text{1 mol O _{2} }}{\text{32.00 g O _{2} }} = \text{2.16 mol O _{2} }$

3. Calculate the moles of CO₂ from each reactant

$\textbf{From C _{8} H _{18} :}\\\text{Moles of CO _{2} } = \text{0.090 17 mol C _{8} H _{18} } \times \dfrac{\text{16 mol CO _{2} }}{\text{2 mol C _{8} H _{18} }} = \text{0.7214 mol CO}_{2}\\\\\textbf{From O _{2} :}\\\text{Moles of CO _{2} } =\text{2.16 molO _{2} } \times \dfrac{\text{16 mol CO _{2} }}{\text{25 mol O _{2} }} = \text{1.38 mol CO _{2} }\\\\\text{Octane is the limiting reactant because it gives fewer moles of CO _{2} .}$

4. Calculate the mass of CO₂

$\text{ Mass of CO _{2} } = \text{0.7214 mol CO _{2} } \times \dfrac{\text{44.01 g CO _{2} }}{\text{1 mol CO _{2} }} = \textbf{31.8 g CO}_\mathbf{{2}}\\\\\text{The reaction produces \large \boxed{\textbf{31.8 g CO}_\mathbf{{2}}} }$