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(Chemistry Ch-1) 6. Stoichiometric Calculations in Balanced Chemical Equations


  • An example of a balanced chemical equation is given below.
From the above balanced chemical equation, the following information is obtained:
  • One mole of C3H8(g) reacts with five moles of O2(g) to give three moles of CO2(g) and four moles of H2O(l).
  • One molecule of C3H8(g) reacts with five molecules of O2(g) to give three molecules of CO2(g) and four molecules of H2O(l).
  • 44 g of C3H8(g) reacts with (5 × 16 = 80) g of O2(g) to give (3 × 44 = 132) g of CO2 and (4 × 18 = 72) g of H2O.
  • 22.4 L of C3H8(g) reacts with (5 × 22.4) L of O2(g) to give (3 × 22.4) L of O2 and (4 × 22.4) L of H2O.
Example
Nitric acid (HNO3) is commercially manufactured by reacting nitrogen dioxide (NO2) with water (H2O). The balanced chemical equation is represented as follows:
Calculate the mass of NO2 required for producing 5 moles of HNO3.
Solution:
According to the given balanced chemical equation, 3 moles of NO2 will produce 2 moles of HNO3.
Therefore, 2 moles of HNO3 require 3 moles of NO2.
Hence, 5 moles of HNO3 require moles of NO2
= 7.5 moles of NO2
Molar mass of NO2 = (14 + 2 × 16) g mol−1
= 46 g mol−1
Thus, required mass of NO2 = (7.5 × 46) g mol−1
= 345 g mol−1
  • Limiting reagent or limiting reactant:
  • Reactant which gets completely consumed when a reaction goes to completion
  • So called because its concentration limits the amount of the product formed


Example
Lead nitrate reacts with sodium iodide to give lead iodide and sodium nitrate in the following manner:
What amount of sodium nitrate is obtained when 30 g of lead nitrate reacts with 30g of sodium iodide?
Solution:
Molar mass of 
= 331 g mol−1
Molar mass of NaI = (23 + 127) = 150 g mol−1
According to the given equation, 1 mole of Pb(NO3)2 reacts with 2 moles of NaI, i.e.,
331 g of Pb(NO3)2 reacts with 300 g of NaI to give PbI2 and NaNO3
Thus, Pb(NO3)2 is the limiting reagent.
Therefore, 30 g of Pb (NO3)2  mole
According to the equation, 0.09 mole of Pb(NO3)2 will give (2 × 0.09) mole of NaNO3 = 0.18 mole of NaNO3.
  • Reactions in solutions:
Ways for expressing the concentration of a solution −
  • Mass per cent or weight per cent (w/w%)
Mass per cent 
Example
4.4 g of oxalic acid is dissolved in 200 mL of a solution. What is the mass per cent of oxalic acid in the solution? (Density of the solution = 1.1 g mL−1)
Solution:
Density of the solution = 1.1 g mL−1
So the mass of the solution = (200 mL) × (1.1 g mL−1)
= 220 g
Mass of oxalic acid = 4.4 g
Therefore, mass per cent of oxalic acid in the solution
  • Mole fraction:
If a substance ‘A’ dissolves in a substance ‘B’, then mole fraction of A 
Mole fraction of B 
nA − Number of moles of A
nB − Number of moles of B
Example
A solution is prepared by dissolving 45 g of a substance X (molar mass = 25 g mol−1) in 235 g of a substance Y (molar mass = 18 g mol−1). Calculate the mole fractions of X and Y.
Solution:
Moles of X, nX = 
= 1.8 mol
Moles of Y, nY = 
= 13.06 mol
Therefore, mole fraction of X, nX 
And, mole fraction of Y, nY = 1 − nX
= 1 − 0.121
= 0.879
  • Molarity:
Number of moles of a solute in 1 L of a solution
Molarity (M) =
Molarity equation:
M1V1 = M2V2
M1 = Molarity of a solution when its volume is V1
M2 = Molarity of the same solution when its volume is V2
Examples
1. 10g of HCl is dissolved in enough water to form 500 mL of the solution. Calculate the molarity of the solution.
Solution:
Molar mass of HCl = 36.5 g mol−1
So the moles of HCl = mol
= 0.274 mol
Volume of the solution = 500 mL = 0.5 L
Therefore, molarity = 
= 0.548 M
2. Commercially available concentrated HCl contains 38% HCl by mass. What volume of concentrated HCl is required to make 2.5 L of 0.2 M HCl? (Density of the solution = 1.19 g mL−1)
Solution:
38% HCl by mass means that 38g of HCl is present in 100 g of the solution.
Moles of HCl = 
Volume of the solution
= 84.03 mL
= 0.08403L
Therefore, molarity of the solution =
= 12.38 M
According to molarity equation,
M1V1 = M2V2
Here,
M1 = 12.38 M
M2 = 0.2 M
V2 = 2.5 L
Now, M1V1 = M2V2
Hence, required volume of HCl = 0.0404 L
  • Molality:
Number of moles of solute present in 1 kg of solvent
Molality (m) = 
Example
What is the molality of a solution of glucose in water, which is labelled as 15% (w/w)?
Solution:
15% (w/w) solution means that 15 g of glucose is present in 100 g of the solution, i.e., (100 − 15) g = 85 g of water = 0.085 kg of water
Moles of glucose = 
= 0.083 mol
Therefore, molality of the solution 
= 0.976 m

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