how many grams of CH4 (methane) must be burned to form 82.6g of H2O?

equation: CH4+O2----> CO2+H2O

See your post above.

To determine the number of grams of CH4 (methane) that must be burned to form 82.6g of H2O, we need to use the stoichiometry of the balanced chemical equation provided.

The balanced equation is: CH4 + 2O2 → CO2 + 2H2O

From the equation, we can see that for every 1 mole of CH4 burned, we obtain 2 moles of H2O. To calculate the amount of CH4 needed, we can follow these steps:

Step 1: Calculate the number of moles of H2O.
Using the molar mass of water (H2O), which is approximately 18 g/mol, we can determine the number of moles of H2O:
Number of moles = mass of H2O / molar mass of H2O
Number of moles = 82.6 g / 18 g/mol ≈ 4.59 moles

Step 2: Apply the stoichiometry from the balanced equation.
Since the ratio of CH4 to H2O is 1:2, the moles of CH4 will also be 2 times the moles of H2O.
Moles of CH4 = 2 × moles of H2O
Moles of CH4 = 2 × 4.59 moles ≈ 9.18 moles

Step 3: Calculate the mass of CH4.
Finally, we can calculate the mass of CH4 by multiplying the number of moles by the molar mass of CH4:
Mass of CH4 = moles of CH4 × molar mass of CH4
Mass of CH4 = 9.18 moles × 16.04 g/mol ≈ 147.25 g

Therefore, approximately 147.25 grams of CH4 must be burned to form 82.6 grams of H2O.