Part A How many grams of \( \mathrm{CO}_{2} \) are produced from 125 g of \( \mathrm{O}_{2} \) and excess \( \mathrm{CH}_{4} \) ? \( \mathrm{CH}_{4}+2 \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O} \) 85.9 g of \( \mathrm{CO}_{2} \) 62.5 g of \( \mathrm{CO}_{2} \) \( 172 \mathrm{~g} \mathrm{of} \mathrm{CO}_{2} \) \( 250 . \mathrm{g} \) of \( \mathrm{CO}_{2} \) 125 g of \( \mathrm{CO}_{2} \) Submit Bequest Answor

To determine how many grams of \( \mathrm{CO}_{2} \) are produced from 125 g of \( \mathrm{O}_{2} \) with excess \( \mathrm{CH}_{4} \), follow these steps: **1. Write the Balanced Chemical Equation:** \[ \mathrm{CH}_{4} + 2\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} + 2\mathrm{H}_{2}\mathrm{O} \] **2. Calculate Molar Masses:** - \( \mathrm{O}_{2} \): \( 32.00 \, \text{g/mol} \) - \( \mathrm{CO}_{2} \): \( 44.01 \, \text{g/mol} \) **3. Convert Grams of \( \mathrm{O}_{2} \) to Moles:** \[ \text{Moles of } \mathrm{O}_{2} = \frac{125 \, \text{g}}{32.00 \, \text{g/mol}} \approx 3.906 \, \text{mol} \] **4. Use Stoichiometry to Find Moles of \( \mathrm{CO}_{2} \):** From the balanced equation, 2 moles of \( \mathrm{O}_{2} \) produce 1 mole of \( \mathrm{CO}_{2} \): \[ \text{Moles of } \mathrm{CO}_{2} = \frac{3.906 \, \text{mol } \mathrm{O}_{2}}{2} \approx 1.953 \, \text{mol} \] **5. Convert Moles of \( \mathrm{CO}_{2} \) to Grams:** \[ \text{Mass of } \mathrm{CO}_{2} = 1.953 \, \text{mol} \times 44.01 \, \text{g/mol} \approx 85.9 \, \text{g} \] **Final Answer:** \[ \boxed{85.9 \, \text{g of } \mathrm{CO}_{2}} \]