Consider the following unbalanced combustion reaction:(a) Balance the equation. (b) How many moles of \(\mathrm{CO}_{2}\) will be generated from using excess methane and \(5.0\) moles of \(\mathrm{O}_{2} ?\) (c) How many molecules of \(\mathrm{CO}_{2}\) will be produced from using excess methane and \(5.0 \mathrm{~g}\) of \(\mathrm{O}_{2}\) ?

To balance the equation, we need to match the number of atoms of each element on both sides of the equation. The balanced equation looks like this: $$\mathrm{CH}_{4} + 2\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} + 2\mathrm{H}_{2}\mathrm{O}$$

Now we calculate how many moles of CO₂ will be generated from using excess methane and \(5.0\) moles of \(\mathrm{O}_{2}\). From the balanced equation, we can see that 1 mole of CH₄ reacts with 2 moles of O₂ to produce 1 mole of CO₂. Based on stoichiometry, $$\frac{\text{moles of CO}_{2}}{\text{moles of O}_{2}}=\frac{1}{2}$$ Let's call the moles of CO₂ produced x. Hence, $$x = \frac{1}{2}\times 5.0\text{ moles of O}_{2}$$ Solve for x: $$x = 2.5\text { moles of CO}_{2}$$ So, 2.5 moles of CO₂ will be generated.

Now we need to determine how many molecules of CO₂ are produced from using excess methane and \(5.0\text{ g}\) of \(\mathrm{O}_{2}\). First, we need to find the moles of \(\mathrm{O}_{2}\) present in \(5.0\text{ g}\) of \(\mathrm{O}_{2}\). We will use the molar mass of \(\mathrm{O}_{2}\) which is \(32.00\text{ g/mol}\), $$\text{moles of O}_{2} = \frac{\text{grams of O}_{2}}{\text{molar mass of O}_{2}}$$ $$\text{moles of O}_{2} = \frac{5.0\text{ g}}{32.00\text{ g/mol}}$$ $$\text{moles of O}_{2} = 0.15625\text{ mol}$$ Now, we calculate the moles of CO₂ produced using the same stoichiometry as in Step 2: $$\text{moles of CO}_{2} = \frac{1}{2}\times 0.15625\text{ moles of O}_{2}$$ $$\text{moles of CO}_{2} = 0.078125\text{ mol}$$ To find the number of molecules of CO₂, we use Avogadro's number (6.022 x 10²³ molecules/mol): $$\text{molecules of CO}_{2} = \text{moles of CO}_{2} \times \text{Avogadro's number}$$ $$\text{molecules of CO}_{2} = 0.078125\text{ mol} \times 6.022\times10^{23}\text{ molecules/mol}$$ $$\text{molecules of CO}_{2} ≈ 4.7\times10^{22}\text{ molecules}$$ So, there are approximately \(4.7\times10^{22}\) molecules of CO₂ produced from using excess methane and \(5.0\text{ g}\) of \(\mathrm{O}_{2}\).