For #CH_4 +2O_2 -> CO_2 + 2H_2O#, the molar mass of oxygen gas (#0_2#) is 32.00 g/mol. The molar mass of carbon dioxide (#CO_2#) is 44.01 g/mol. What mass of #CO_2#, in grams, will form when 8.94 g #O_2# completely react?

You must calculate the molar quantity of the given mass of dioxygen and with how much methane it will react. The balanced chemical equation clearly states that 16 g of methane gas reacts with 32 g of dioxygen gas to give 44 g of carbon dioxide and 36 g of water. These masses represent the molar equivalence of reactants and products.

22.05 grams of CO₂ will form when 8.94 grams of O₂ completely react.

To find the mass of CO2 formed when 8.94 g of O2 completely reacts, we need to use stoichiometry to relate the masses of the reactants and products.

1. Start by calculating the moles of O2 using its molar mass: ( \text{Molar mass of O}_2 = 32.00 , \text{g/mol} ) ( \text{Moles of O}_2 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{8.94 , \text{g}}{32.00 , \text{g/mol}} )

2. Use the balanced chemical equation to relate the moles of O2 to the moles of CO2: ( \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} ) From the equation, we see that 1 mole of CH4 reacts with 2 moles of O2 to produce 1 mole of CO2. So, the mole ratio of O2 to CO2 is 2:1.

3. Calculate the moles of CO2 using the mole ratio: Moles of CO2 = ( \frac{2}{1} \times \text{Moles of O}_2 )

4. Finally, find the mass of CO2 formed using its molar mass: Mass of CO2 = Moles of CO2 × Molar mass of CO2

Let's plug in the values:

[ \text{Moles of O}_2 = \frac{8.94 , \text{g}}{32.00 , \text{g/mol}} ] [ \text{Moles of CO}_2 = \frac{2}{1} \times \text{Moles of O}_2 ] [ \text{Mass of CO}_2 = \text{Moles of CO}_2 \times \text{Molar mass of CO}_2 ]

After calculating these values, you will find the mass of CO2 formed when 8.94 g of O2 completely reacts.