how many grams of CH4 (methane) must be burned to form 82.6g of H2O?
equation: CH4+O2----> CO2+H2O
CH4 + 2O2----> CO2 + 2H2O
mols H2O = grams/molar mass H2O
Using the coefficients in the balanced equation, convert mols H2O to mols CH4.
Now convert mols CH4 to g. grams = mols x molar mass CH4
To find the number of grams of CH4 (methane) needed to form 82.6g of H2O, you need to use stoichiometry, which allows you to relate the amounts of substances involved in a chemical reaction.
First, you need to determine the molar ratios between CH4 and H2O using the balanced chemical equation:
CH4 + 2O2 → CO2 + 2H2O
From the balanced equation, you can see that 1 mole of CH4 reacts with 2 moles of H2O.
Next, convert the given mass of H2O into moles. To do this, you need the molar mass of water (H2O), which is 18.015 g/mol. Divide the given mass (82.6g) by the molar mass to obtain moles:
82.6g H2O / 18.015 g/mol = 4.58 moles H2O
Finally, using the molar ratio from the balanced equation, you can determine the moles of CH4 needed to produce the given amount of H2O. Since the ratio is 1:2 (CH4:H2O), the moles of CH4 can be calculated as:
4.58 moles H2O × (1 mole CH4 / 2 moles H2O) = 2.29 moles CH4
Now, you can find the mass of CH4 in grams by multiplying the moles by the molar mass of CH4, which is 16.04 g/mol:
2.29 moles CH4 × 16.04 g/mol = approximately 36.7 grams of CH4
Therefore, you would need approximately 36.7 grams of CH4 to form 82.6 grams of H2O.