Courses

# Stoichiometry and Stoichiometric Calculations Class 11 Notes | EduRev

## JEE : Stoichiometry and Stoichiometric Calculations Class 11 Notes | EduRev

The document Stoichiometry and Stoichiometric Calculations Class 11 Notes | EduRev is a part of the JEE Course Chemistry Class 11.
All you need of JEE at this link: JEE

STOICHIOMETRY
Stoichiometry is the calculations of the quantities of reactants and products involved in a chemical reaction. Following methods can be used for solving problems.
(a) Mole Method (For Balance reaction)
(b) POAC method (Balancing not required but common sense, use it with slight care)
(c) Equivalent concept

CONCEPT OF LIMITING REAGENT
Limiting Reagent: It is very important concept in chemical calculation. It refers to reactant which is present in minimum stoichiometry quantity for a chemical reaction. Fig. Concept of Limiting reagent.

It is the reactant consumed fully in a chemical reaction. So all calculations related to various products or in sequence of reactions are made on the basis of limiting reagent.
It comes into picture when reaction involves two or more reactants. For solving any such reaction, first step is to calculate L.R.

Calculation of Limiting Reagent
(a) By calculating the required amount by the equation and comparing it with given amount.
[Useful when only two reactants are there].

(b) By calculating amount of any one product obtained taking each reactant one by one irrespective of other reactants. The one giving least product is limiting reagent.

(c) Divide given moles of each reactant by their stoichiometric coefficient, the one with least ratio is limiting reagent. [Useful when the number of reactants are more than two.]

PERCENTAGE YIELD
The percentage yield of product The actual amount of any limiting reagent consumed in such incomplete reactions is given by:
% yield × given moles of limiting reagent [For reversible reactions].

Example. A compound which contains one atom of X and two atoms of Y for each three atoms of Z is made by mixing 5 gm of X, 1.15 × 1023 atoms of Y and 0.03 mole of Z atoms. Given that only 4.40 gm of compound results. Calculate the atomic weight of Y if atomic weight of X and Z are 60 and 80 respectively.
Solution.
Moles of x = 5/60 = 1/12 = 0.083 Moles of z = 0.03
x + 2y + 3z → xy2z3
For limiting reagent, 0.083/1 = 0.083
0.19/2 = 0.095 , 0.03/3 = 0.01
Hence z is limiting reagent
wt of xy2z3 = 4.4 gm = moles × molecular wt.
Moles of xy2z3 = 1/3 × 0.03 = 0.01
300 + 2m = 440
⇒ 2m = 440 - 300 ⇒ m =70

POAC Rule
POAC is the simple mass conservation:
KClO3 → KClO2
Apply the POAC on K.
moles of K in KClO3 = moles of K in KCl
1 × moles of KClO3 = 1 × moles of KCl
moles of KClO3 = moles of KCl
Apply POAC on O
moles O in KClO= moles of O in O2
3 × moles of KClO3 = 2 × moles of O2 Fig. Structure of KClO3

Example 1. In the gravimetric determination of phosphorous, an aqueous solution of dihydrogen phosphate ion (H2PO4-) is treated with a mix of ammonium & magnesium ions to precipitate magnesium ammonium phosphate MgNH4 PO4.6H2O. This is heated and decomposed to magnesium Pyrophosphate, Mg2P2O7 which is weighted. A solution of H2PO4- yielded 1.054 gm of Mg2P2O7 what weight of NaH2PO4 was present originally.
NaH2PO4 → Mg2P2Oapply POAC on P
Solution.
Let wt of NaH2PO4 = w gm
moles of P in NaH2PO4 = moles of P in Mg2P2O7
w/120 × 1 = 1.054/232 × 2
w = 1.054 x 120/232 × 2 = 1.09 gm

SOME EXPERIMENTAL METHODS
For determination of atomic mass
Dulong's and Petit's Law:
Atomic weight × specific heat (cal/gm°C) ∝ ≌ 6.4
Gives approximate atomic weight and is applicable for metals only.
Take care of units of specific heat.

Example 2. 7.5 mL of a hydrocarbon gas was exploded with excess of oxygen. On cooling, it was found to have undergone a contraction of 15 mL. If the vapour density of the hydrocarbon is 14, determine its molecular formula. (C = 12, H = 1)
Solution. on cooling the volume contraction = 15 ml
i.e The volume of H2O (g) = 15 ml
V.D of hydrocarbon = 14
Molecular wt. of CxHx = 28
12x + y = 28 ...(1)
From reaction 12 x + 4 = 28
12 x = 24
x = 2
Hence Hydrocarbon is C2H4.

Example 3. Calculate the weight of FeO produced from 2g VO and 5.75 g of Fe2O3. Also report the limiting reagent.
VO + Fe2O3 → FeO + V2O5
Solution.
Balancing the given equation
2VO + Fe2O3 → 6FeO + V2O5
Mole before reaction = 2/67 = 5.75/160
0.0298 = 0.0359 = 0 = 0
Mole after reaction Mole of FeO formed = 0.0359 × 2
Weight of FeO formed = 0.0359 × 2 × 72
= 5.17 g
The limiting reagent is one which is used completely, i.e. Fe2O3 here.

Try Yourself!
Q.1. What is the number of moles of Fe(OH)3 (s) that can be produced by allowing 1 mole of Fe2S3, 2 moles of H2O and 3 moles of O2 to react ?
2Fe2S3 (s) + 6H2O(l) + 3O2(g ) → 4Fe (OH )3(s) + 6S(s)
Ans. 1.34 moles

Q.2. In a process for producing acetic acid, oxygen gas is bubbled into acetaldehyde containing manganese (II) acetate (catalyst) under pressure at 60ºC.
2CH3CHO + O2 →2CH3COOH
In a laboratory test of this reaction, 20 g of CH3CHO and 10 g of O2 were put into a reaction vessel.
(a) How many grams of CH3COOH can be produced ?
(b) How many grams of the excess reactant remain after the reaction is complete?
Ans.
(a) 27.27 g
(b) 2.73 g

Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

## Chemistry Class 11

136 videos|245 docs|180 tests

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

;