When 10.0 grams of ch4 reacts completely with 40.0 grams of o2 such that there are no reactants left over, 27.5 grams of carbon dioxide are formed. how many grams of water are formed? ch4+ 2o2 â co2 + 2h2o?

The balanced equation for the reaction is :-

CH₄(g) + 2O₂(g)  > CO₂(g) + 2H₂O(l)

Molar mass of CH₄ = 16 g/mole

Molar mass of O₂ = 32 g/mole

Molar mass of H₂O = 18 g/mole

Molar mass of CO₂ = 44 g/mole

Now we calculate the number of moles of reactants,

moles of CH₄ = mass of CH₄/molar mass of CH₄ = 10/16 = 0.625

Moles of O₂ = mass of O₂/molar mass of O₂ = 40/32 = 1.25

Now, as per the balanced reaction, for complete reaction to occur, one mole of CH₄ require 2 moles of O₂

Thus, 0.625 moles of CH₄ requires 1.25 moles of O₂

both CH₄ and O₂ are present in the exact required quantity.

Hence, moles of CO₂ formed = moles of CH₄ reacted = 0.625

Mass of CO₂ formed = moles of CO₂ x Molar mass of CO₂ = 0.625 x 44 = 27.5g

Moles of H₂O formed = moles of O₂ reacted = 1.25g

Thus, mass of H₂O formed = moles of H₂O x molar mass of H₂O = 1.25 x 18 = 22.5g

the answer is a! (2.9 x 10^-18 j) hope i could someone. : )