Question 6.1
Question 6.1.1 [2 marks]
Define limiting reagent.
The limiting reagent (or limiting reactant) is the reactant in a chemical reaction that is completely consumed first, limiting the amount of product that can be formed. Once the limiting reagent is used up, the reaction stops, even if other reactants are still available.
Question 6.1.2 [6 marks]
Determine which reagent is in excess.
First, we need to convert the masses of the reactants to moles using their molar masses.
- Molar mass of \( \text{H}_2\text{SO}_4 \) = 2(1) + 32 + 4(16) = 98 g/mol
- Molar mass of \( \text{NH}_3 \) = 14 + 3(1) = 17 g/mol
Now, we can calculate moles of each reactant:
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For \( \text{H}_2\text{SO}_4 \): \[ \text{Mass of } \text{H}_2\text{SO}_4 = 3 , \text{kg} = 3000 , \text{g} \] \[ \text{Moles of } \text{H}_2\text{SO}_4 = \frac{3000 , \text{g}}{98 , \text{g/mol}} \approx 30.61 , \text{mol} \]
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For \( \text{NH}_3 \): \[ \text{Mass of } \text{NH}_3 = 1.2 , \text{kg} = 1200 , \text{g} \] \[ \text{Moles of } \text{NH}_3 = \frac{1200 , \text{g}}{17 , \text{g/mol}} \approx 70.59 , \text{mol} \]
In the balanced equation: \[ \text{H}_2\text{SO}_4 + 2\text{NH}_3 \rightarrow (\text{NH}_4)_2\text{SO}_4 \]
From the balanced equation, 1 mole of \( \text{H}_2\text{SO}_4 \) reacts with 2 moles of \( \text{NH}_3 \). Therefore, for every mole of \( \text{H}_2\text{SO}_4 \), we need 2 moles of \( \text{NH}_3 \).
Now calculate the required moles of \( \text{NH}_3 \) for the available moles of \( \text{H}_2\text{SO}_4 \): \[ \text{Required } \text{NH}_3 = 30.61 , \text{mol} \times 2 = 61.22 , \text{mol} \]
Since we have approximately 70.59 moles of \( \text{NH}_3 \), which is more than the required amount (61.22 mol), and only 30.61 moles of \( \text{H}_2\text{SO}_4 \), \( \text{H}_2\text{SO}_4 \) is the limiting reagent.
Excess reagent is \( \text{NH}_3 \).
Question 6.1.3 [2 marks]
Calculate the maximum amount of ammonium sulphate produced in kg.
From the balanced equation, 1 mole of \( \text{H}_2\text{SO}_4 \) yields 1 mole of \( (\text{NH}_4)_2\text{SO}_4 \).
- Molar mass of \( (\text{NH}_4)_2\text{SO}_4 \): \[ = 2(14) + 8(1) + 32 + 4(16) = 132 , \text{g/mol} \]
Since \( \text{H}_2\text{SO}_4 \) is the limiting reagent (30.61 mol), the same number of moles of ammonium sulfate will be produced.
Now, calculate the mass of ammonium sulfate produced: \[ \text{Mass of } (\text{NH}_4)_2\text{SO}_4 = 30.61 , \text{mol} \times 132 , \text{g/mol} = 4040.52 , \text{g} \approx 4.04 , \text{kg} \]
Question 6.2
Question 6.2.1 [2 marks]
Define empirical formula.
The empirical formula of a compound is the simplest positive integer ratio of the atoms of each element present in the compound.
Question 6.2.2 [5 marks]
Calculate the empirical formula of the compound.
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Convert percentages to grams assuming 100 g of the substance:
- K: 55.3 g
- P: 14.6 g
- O: 30.1 g
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Convert grams to moles:
- Moles of K: \[ \frac{55.3 , \text{g}}{39.1 , \text{g/mol}} \approx 1.414 , \text{mol} \]
- Moles of P: \[ \frac{14.6 , \text{g}}{30.97 , \text{g/mol}} \approx 0.471 , \text{mol} \]
- Moles of O: \[ \frac{30.1 , \text{g}}{16.00 , \text{g/mol}} \approx 1.88 , \text{mol} \]
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Find the simplest mole ratio:
- Divide each by the smallest number of moles (0.471):
- K: \( \frac{1.414}{0.471} \approx 3 \)
- P: \( \frac{0.471}{0.471} \approx 1 \)
- O: \( \frac{1.88}{0.471} \approx 4 \)
- Divide each by the smallest number of moles (0.471):
Thus, the empirical formula is \( \text{K}_3\text{P}\text{O}_4 \).
Question 6.2.3 [3 marks]
Calculate the molecular formula of the compound.
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Calculate the molar mass of the empirical formula \( \text{K}_3\text{P}\text{O}_4 \):
- Molar mass of \( K = 39.1 \times 3 = 117.3 , \text{g/mol} \)
- Molar mass of \( P = 30.97 , \text{g/mol} \)
- Molar mass of \( O = 16.00 \times 4 = 64.00 , \text{g/mol} \)
Total molar mass of \( \text{K}_3\text{P}\text{O}_4 \): \[ 117.3 + 30.97 + 64.00 \approx 212.27 , \text{g/mol} \]
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Calculate the ratio of the molar mass of the compound to the empirical formula's molar mass: \[ \frac{212 , \text{g/mol}}{212.27 , \text{g/mol}} \approx 1 \]
Since the ratio is 1, the molecular formula is the same as the empirical formula: \[ \text{Molecular Formula} = \text{K}_3\text{P}\text{O}_4 \]
This concludes the solution to the questions posed.
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