Quick ! during a combustion reaction, 5.00 grams of oxygen reacted with 5.00 grams of ch4. what is the amount of the leftover reactant? 4.00 grams of methane 3.75 grams of methane 2.75 grams of oxygen 1.75 grams of oxygen

The balanced chemical reaction is:

CH4 + 2O2 —> CO2 + 2H2O

You need to convert mass to moles (divide by molar mass):

CH4 moles = 5 / 16 = 0.31 mol
O2 moles = 5 / 32 = 0.16 mol

To figure out which reactant is limiting, divide the actual moles by the corresponding coefficient in the reaction:

CH4: 0.31 / 1 = 0.31
O2: 0.16 / 2 = 0.08

O2 is the lower number, so it is the limiting reactant. From the reaction we know it takes 2 moles of O2 to react with each mole of CH4. Therefore, for however many moles of O2 we actually have, half as many moles of CH4 will react. Since we have 0.16 mol of O2, only 0.08 mol of CH4 will react, leaving behind 0.31 - 0.08 = 0.23 mol of CH4.

Now convert back to mass (multiply by molar mass) to find the mass of CH4 remaining:

0.23 x 16 = 3.68g

The closest answer is B.

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