Chemistry: Stoirchiometry

Question

(3.5000x10^-1) L of methane, CH₄(g), is combusted with (1.40x10^-1) L of oxygen gas in SATP conditions. Once the reaction is completely finished, how many moles of the excess reagent remains?

Note: Your answer is assumed to be reduced to the highest power possible and sig digs.

Answer 1

Looking at the balanced equation for the reaction taking place:

CH₄(g) + 2O₂(g) ==> CO₂(g) + 2H₂O(g)

SATP = Standard Ambient Temperature and Pressure: T=25ºC = 298K; P=100 kPa;

molar volume = 24.5 L/mole

moles CH4 present = 3.5 L x 1 mol / 24.5 L = 0.1429 mols CH4

moles O2 present = 1.40x10-1 L = 0.140 L x 1 mol / 24.5 L = 0.0057 mols O2

Limiting reactant = O2(g)

moles CH4 used up = 0.0057 mols O2 x 1 mol CH4 / 2 mol O2 = 0.0029 mols CH4 used

moles CH4 remaining = 0.1429 mols - 0.0029 mols = 0.140 mols left over (3 sig. figs.)

Once again, I have no idea what is meant by reducing your answer to the highest power.

Would that be 140x10^-3 moles

or 0.140x10⁰ moles?

or 0.000000000140x10⁹ moles? Etc.....