Question 3.1
Question 3.1.1 [1 mark]
Define a limiting reagent. A limiting reagent is the reactant that is completely consumed in a chemical reaction, thus determining the maximum amount of product that can be formed. Once this reactant is used up, the reaction cannot proceed further, even if other reactants are still available.
Question 3.1.2 [6 marks]
Determine which of the two reactants is in excess.
To determine the limiting reagent, we need to calculate the number of moles of each reactant:
-
Calculate moles of hydrogen (H2): \[ \text{Molar mass of } H_2 = 2.02 , \text{g/mol} \] \[ \text{Moles of } H_2 = \frac{2.50 , \text{g}}{2.02 , \text{g/mol}} \approx 1.237 , \text{mol} \]
-
Calculate moles of carbon monoxide (CO): \[ \text{At STP, } 1 , \text{mol of gas occupies } 22.4 , \text{dm}^3. \] \[ \text{Moles of } CO = \frac{30.0 , \text{dm}^3}{22.4 , \text{dm}^3/\text{mol}} \approx 1.339 , \text{mol} \]
-
Stoichiometry of the reaction: The balanced equation is: \[ CO(g) + 2H_2(ℓ) \rightarrow CH_3OH(l) \] From the equation, 1 mole of CO reacts with 2 moles of H2. Therefore, for 1.339 moles of CO, it would require: \[ \text{Required moles of } H_2 = 2 \times 1.339 = 2.678 , \text{mol} \]
Since we only have approximately 1.237 moles of H2, hydrogen is the limiting reagent, and CO is in excess.
Question 3.1.3 [3 marks]
Calculate the theoretical mass of the product formed.
-
From the balanced reaction equation, 2 moles of H2 produce 1 mole of CH3OH.
-
Moles of CH3OH produced from 1.237 moles of H2: \[ \text{Moles of } CH_3OH = \frac{1.237}{2} = 0.6185 , \text{mol} \]
-
Calculate the mass of CH3OH produced: \[ \text{Molar mass of } CH_3OH = 12.01 + (3 \times 1.01) + 16.00 \approx 32.04 , \text{g/mol} \] \[ \text{Mass of } CH_3OH = 0.6185 , \text{mol} \times 32.04 , \text{g/mol} \approx 19.8 , \text{g} \]
Question 3.1.4 [2 marks]
If the actual yield of the product is 15g, calculate the percentage yield of the reaction.
\[ \text{Percentage yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100 = \left( \frac{15 , \text{g}}{19.8 , \text{g}} \right) \times 100 \approx 75.76% \]
Question 3.2 [6 marks]
An impure sample KO2 with a mass of 150.0g is reacted to produce 89.7g of K2CO3. What is the percentage purity of KO2?
-
Balanced reaction: \[ 4KO_2(s) + 2CO_2(g) \rightarrow 2K_2CO_3(s) + 3O_2(g) \]
-
Molar mass of K2CO3: \[ \text{Molar mass of } K_2CO_3 = 2(39.10) + 12.01 + 3(16.00) \approx 138.21 , \text{g/mol} \]
-
Moles of K2CO3 produced: \[ \text{Moles of } K_2CO_3 = \frac{89.7 , \text{g}}{138.21 , \text{g/mol}} \approx 0.649 , \text{mol} \]
-
Calculate moles of KO2 reacted: From the balanced equation, \(2 , \text{mol K2CO3} \) is produced from \(4 , \text{mol KO2}\): \[ \text{Moles of } KO_2 = 4 \times \frac{0.649}{2} = 1.298 , \text{mol} \]
-
Mass of KO2 that reacted: Molar mass of KO2: \[ \text{Molar mass of } KO_2 = 39.10 + 16.00 \times 2 \approx 71.10 , \text{g/mol} \] \[ \text{Mass of } KO_2 = 1.298 , \text{mol} \times 71.10 , \text{g/mol} \approx 92.16 , \text{g} \]
-
Percentage purity of KO2: \[ \text{Percentage Purity} = \left( \frac{92.16 , \text{g}}{150.0 , \text{g}} \right) \times 100 \approx 61.44% \]
Question 3.3.1 [1 mark]
Define Empirical formula. The empirical formula is the simplest, integer ratio of the atoms of each element present in a compound, representing the lowest whole number ratio of the elements, rather than the actual number of atoms in the molecule.
1 answer