(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 C₃H_8(g) reacts with five moles of O_2(g) to give three moles of CO_2(g) and four moles of H₂O_(l).
• One molecule of C₃H_8(g) reacts with five molecules of O_2(g) to give three molecules of CO_2(g) and four molecules of H₂O_(l).
• 44 g of C₃H_8(g) reacts with (5 × 16 = 80) g of O_2(g) to give (3 × 44 = 132) g of CO₂ and (4 × 18 = 72) g of H₂O.
• 22.4 L of C₃H_8(g) reacts with (5 × 22.4) L of O_2(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 

n_A − Number of moles of A

n_B − 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