I combusted methane and got three moles of water. How many grams of methane reacted with excess oxygen to produce that much water?
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Answer:
24g
Explanation:
First, the balanced equation for the reaction.
CH4 + 2O2 —> CO2 + 2H2O
Next, we shall determine the number of mole of CH4 that produced 3 moles of H2O.
This is illustrated below:
From the balanced equation above,
1 mole of CH4 reacted to produce 2 moles of H2O.
Therefore, Xmol of CH4 will react to produce 3 moles of H2O i.e
Xmol of CH4 = 3/2
Xmol of CH4 = 1.5 mole
Finally, we shall convert 1.5 mole of CH4 to grams.
This is shown below:
Number of mole of CH4 = 1.5 moles
Molar mass of CH4 = 12 + (4x1) = 16g/mol
Mass of CH4 =..?
Mass = number of mole x molar Mass
Mass of ch4= 1.5 x 16
Mass of CH4 = 24g