Limiting and excess reagent

WHAT IS LIMITING AND EXCESS REAGENT

Limiting reactant problems involves determining from the information given the inital quanities if the reactants in moles and then determining which one(s) are in excess and which reactant is the limiting reactant. Remember mole is KING.
An example of a limiting and excess reagent is a bike.


An image of a bike with 2 wheels and 1 frame.
The equation here is 2 wheels + 1 frame = 1 Bike
It is also called a 2:1 ratio.

Heres an question,
19g of C7H16 reacts with 113g of O2. Determine the mole of H2O in the following reaction.
C7H16 + 11O2 -> 7CO2 + 8H2O (balanced equation)
Must calculate the moles of C7H16 and O2.

n (C7H16)=m/Mr
n (C7H16)= 19g/100g/mol
n (C7H16)=0.19 mol

n (O2)=m/Mr
n (O2)=113g/32g/mol
n (O2)=3.35 mol

Next step is to determine which of the reactants is the limiting and the excess.
0.19 mol of C7H16 wants to react with 11/1 *0.19 mol = 2.09 mol of oxygen.
0.19 mole of C7H16 wants 2.01. Since 2.01 mol of oxygen is less than what i have (3.35 mol) it means oxygen is in EXCESS and C7H16 is the LIMITING. Due to the fact that C7H16 is the limiting, we bring the coefficient of C7H16 which is 1 to the other side of the equation. So to calculate the mole of H2O you place the coefficient of C7H16 under the coefficient of H2O and multiply it by the limiting mole of C6H17 which is 0.19 mol.

C7H16 + 11O2 -> 7CO2 + 8H2O
n (H2O)=8/1 *0.19 mole
n (H2O)=1.52 molw

The mole of H2O is 1.52 mol