This document discusses various concepts related to stoichiometry including limiting reagents, concentration of solutions, mass percent, mole fraction, molarity, and molality. It provides examples of calculating the limiting reagent, mass percent, and mole fraction in chemical reactions and solutions. Key steps include calculating the moles of each reactant, determining which has the lower moles and is therefore limiting, and using mole ratios from a balanced equation to calculate grams of products and excess reactants.

3. It includes- Limiting reagent &
Concentration of solution

4. In the last part we have seen about Stoichiometry and limiting
reagent. The following reaction shows limiting and excess reagent
As in the above reaction the green color
substance is completely used up, it is a limiting
reagent and the red one is still remain unused it
is excess reagent.

5. The limiting reagent must be identified in order to calculate the percentage yield of a reaction
since the theoretical yield is defined as the amount of product obtained when the limiting
reagent reacts completely. Given is the balanced chemical equation, which describes the
reaction, there are several equivalent ways to identify the limiting reagent and evaluate the
excess quantities of other reagents. One of them is as follows;
Just calculate this formula for each reagent, and the reagent that has the
lowest value of this formula is the limiting reagent. we can apply this
shortcut in the above example.

6. Finding the limiting reagent

7. What is the limiting reagent 76.4 grams of C2H3Br3 were
reacted with 49.1 grams of O2?
: Mass of C2H3Br3 = 266.72g

8. It includes:
Mass percent, Mole fraction, Molarity, Molality

9. Mass percentage
The mass percentage is one way of representing the concentration of an element in a
compound or a component in a mixture. The mass percentage is calculated as the mass of a
component divided by the total mass of the mixture, multiplied by 100%.
 The formula for a solution is:
 mass percent = (grams of solute / grams of solute + solvent) x 100
or
 mass percent = (grams of solute / grams of solution) x 100
 The final answer is given as %.
 Also Known As: mass percent, (w/w)%
Definition & Example

11. Find the mass percentage of 6 g sodium hydroxide dissolved
in 50 g of water. (Note: since the density of water is nearly
1, this type of question often gives the volume of water in
milliliters.)
 Solution:
 First find the total mass of the solution:
 total mass = 6 g sodium hydroxide + 50 g water
total mass = 56 g
 Now, you can find the mass percentage of the sodium hydroxide
using the formula:
 mass percent = (grams of solute / grams of solution) x 100
mass percent = (6 g NaOH / 56 g solution) x 100
mass percent = (0.1074) x 100
answer = 10.74% NaOH

12.  Mole fraction
It is the ratio of the number of moles of a particular component to the total
number of moles of all the components in the mixture. It is denoted by symbol χ.
If a substance A of nA moles is dissolves in solution of B having nB ,then
mole fraction is expressed as
denoted by greek letter Χ(chi)
For two components
where, n = moles
moles=Weight/Molar Mass

14. Example of mole fraction
 Example #1: A solution is prepared by mixing 25.0 g of water, H2O,
and 25.0 g of ethanol, C2H5OH. Determine the mole fractions of each
substance.
 Solution: 1) Determine the moles of each substance:
H2O ⇒ 25.0 g / 18.0 g/mol = 1.34 mol
C2H5OH ⇒ 25.0 g / 46.07 g/mol = 0.543 mol
Example# 2:
Find the mole fraction of Methanol CH3OH and water in a solution prepared by dissolving 4.5 g of
alcohol in 40 g of H2O.
Molar Mass of H2O is 18gm/mole and Molar mass of CH3OH is 32gm/mole
Solution:
Moles of CH3OH = 4.5 / 32 = 0.14 mole Moles of H2O = 40 / 18 = 2.2 moles
Therefore, according to the equation
Mole fraction of CH3OH = 0.14 / 2.2 + 0.14 = 0.061, Mole fraction of H2O. = 2.2 / 2.2 + 0.14 = 0.940
2) Determine mole fractions:
H2O ⇒ 1.34 mol / (1.34 mol + 0.543
mol) = 0.71
C2H5OH ⇒ 0.543 mol / (1.34 mol +
0.543 mol) = 0.29

15.  Find the masses of sodium chloride and water required to obtain
175 g of a 15% solution.
 What is the mass percent of hydrogen in water?
 A solution contains 10.0 g pentane, 10.0 g hexane and 10.0 g
benzene. What is the mole fraction of hexane?
 Calculate the mole fraction of HCl and H2O in a solution of
hydrochloric acid in water, containing 30% HCl by weight.
 In a solution , we have 1 mole of A , 1 mole of B and 2 mole of C.
Find the mole fraction of each of them.

16.  A reaction container holds 5.77 g of P4 and 5.77 g of O2.
The following reaction occurs: P4 + O2 → P4O6. If enough oxygen is
available then the P4O6 reacts further: P4O6 + O2 → P4O10.
a. What is the limiting reagent for the formation of P4O10?
b. What mass of P4O10 is produced?
c. What mass of excess reactant is left in the reaction container?
Find out
 Take the reaction: NH3 + O2 → NO + H2O. In an experiment, 3.25 g of
NH3 are allowed to react with 3.50 g of O2.
a. Which reactant is the limiting reagent?
b. How many grams of NO are formed?
c. How much of the excess reactant remains after the reaction?