First, we need to determine the number of moles of methane (CH4) present in 98.0 grams:
98.0 g CH4 / 16 g/mol = 6.125 moles CH4
Next, we need to determine the limiting reactant by calculating the number of moles of oxygen (O2) required to react with 6.125 moles of CH4:
6.125 moles CH4 x (2 moles O2 / 1 mole CH4) = 12.25 moles O2
Now we need to calculate the number of moles of water (H2O) that can be produced from 6.125 moles of CH4:
6.125 moles CH4 x (2 moles H2O / 1 mole CH4) = 12.25 moles H2O
Finally, we can determine the mass of water produced:
12.25 moles H2O x 18 g/mol = 220.5 grams of water
Therefore, the correct answer is a) 220.5 grams of water.
If 98.0 grams of methane (CH4) react completely with oxygen (O2), how many grams of water (H2O) will be produced? The molar mass of methane(CH4) is 16 g/mol. The molar mass of water (H2O) is 18 g/mol. If you are unsure how to convert from grams to grams, please see the example problem linked HERE. a 220.5 grams of water b 144.0 grams of water c 55.1 grams of water d 174.2 grams of Water
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