Answer:
[tex]\large \boxed{\text{83.5 g CO}_{2}}[/tex]
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: 12.01 32.00 44.01
C + O₂ ⟶ CO₂
Mass/g: 22.8 78.9
2. Calculate the initial moles of C
We know that there is excess oxygen, so C is the limiting reactant.
[tex]\text{Moles of C} = \text{22.8 g C} \times \dfrac{\text{1 mol C}}{\text{12.01 g C}} = \text{1.898 mol C}[/tex]
3. Calculate the moles of CO₂ formed
[tex]\text{Moles of CO$_{2}$} = \text{1.898 mol C} \times \dfrac{\text{1 mol CO$_{2}$}}{\text{1 mol C}} = \text{1.898 mol CO}_{2}\\[/tex]
4. Calculate the mass of CO₂
[tex]\text{ Mass of CO$_{2}$} = \text{1.898 mol CO$_{2}$} \times \dfrac{\text{44.01 g CO$_{2}$}}{\text{1 mol CO$_{2}$}} = \textbf{83.5 g CO}_\mathbf{{2}}\\\\\text{The reaction produced $\large \boxed{\textbf{83.5 g CO}_\mathbf{{2}}}$}[/tex]
