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I combusted methane and got three moles of water. How many grams of methane reacted with excess oxygen to produce that much water?

I combusted methane and got three moles of water How many grams of methane reacted with excess oxygen to produce that much water class=

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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