Answer:
0.236 moles of methane gas are required when there are 7.55 grams of oxygen gas available.
Step-by-step explanation:
The balanced reaction is
CH₄ + 2 O₂ → CO₂ + 2 H₂O
By reaction stoichiometry (that is, the ratio of reactants and products in the reaction), the following amounts of moles of each compound participate in the reaction:
- CH₄: 1 mole
- O₂: 2 moles
- CO₂: 1 mole
- H₂O: 2 moles
Being the molar mass of each compound:
- CH₄: 16 g/mole
- O₂: 32 g/mole
- CO₂: 44 g/mole
- H₂O: 18 g/mole
Then, by reaction stoichiometry, the following mass amounts of each compound participate in the reaction:
- CH₄: 1 moles* 16 g/mole= 16 grams
- O₂: 2 moles* 32 g/mole= 64 grams
- CO₂: 1 mole* 44 g/mole= 44 grams
- H₂O: 2 moles* 18 g/mole= 36 grams
So, you can apply the following rule of three: if by reaction stoichiometry 64 grams of oxygen react with 2 moles of methane, 7.55 grams of oxygen react with how many moles of methane?
moles of methane= 0.236
0.236 moles of methane gas are required when there are 7.55 grams of oxygen gas available.