Solveeit Logo
Login

Question

Question: A sample of KClO3 on decomposition yielded 448 mL of oxygen gas at NTP. Calculate (i) Weight of oxyg...

A sample of KClO3 on decomposition yielded 448 mL of oxygen gas at NTP. Calculate (i) Weight of oxygen product, (ii) Weight of KClO3 originally taken and (iii) Weight of KCI produced. (K = 39, Cl = 35.5 and O = 16)

Answer

(i) Weight of oxygen product = 0.64 g (ii) Weight of KClO3_3 originally taken ≈ 1.633 g (iii) Weight of KCl produced ≈ 0.993 g

Explanation

Solution

The decomposition of KClO3_3 is represented by the balanced chemical equation:

2KClO3(s)→Δ2KCl(s)+3O2(g)2KClO_3(s) \xrightarrow{\Delta} 2KCl(s) + 3O_2(g)

Given: Volume of O2O_2 produced at NTP = 448 mL Atomic masses: K = 39, Cl = 35.5, O = 16

At NTP (Normal Temperature and Pressure), 1 mole of any gas occupies 22.4 L or 22400 mL.

  1. Calculate the number of moles of O2O_2 produced: Moles of O2=Volume of O2 (mL)Molar volume at NTP (mL/mol)O_2 = \frac{\text{Volume of } O_2 \text{ (mL)}}{\text{Molar volume at NTP (mL/mol)}} Moles of O2=448 mL22400 mL/mol=0.02 molO_2 = \frac{448 \text{ mL}}{22400 \text{ mL/mol}} = 0.02 \text{ mol}

  2. Calculate the weight of oxygen product: Molar mass of O2=2×16=32 g/molO_2 = 2 \times 16 = 32 \text{ g/mol} Weight of O2=Moles of O2×Molar mass of O2O_2 = \text{Moles of } O_2 \times \text{Molar mass of } O_2 Weight of O2=0.02 mol×32 g/mol=0.64 gO_2 = 0.02 \text{ mol} \times 32 \text{ g/mol} = 0.64 \text{ g}

  3. Calculate the weight of KClO3_3 originally taken: From the balanced equation, 2 moles of KClO3KClO_3 produce 3 moles of O2O_2. The mole ratio of KClO3KClO_3 to O2O_2 is 2:3. Moles of KClO3=Moles of O2×(2 mol KClO33 mol O2)KClO_3 = \text{Moles of } O_2 \times \left(\frac{2 \text{ mol } KClO_3}{3 \text{ mol } O_2}\right) Moles of KClO3=0.02 mol×(23)=0.043 molKClO_3 = 0.02 \text{ mol} \times \left(\frac{2}{3}\right) = \frac{0.04}{3} \text{ mol} Molar mass of KClO3=39+35.5+(3×16)=39+35.5+48=122.5 g/molKClO_3 = 39 + 35.5 + (3 \times 16) = 39 + 35.5 + 48 = 122.5 \text{ g/mol} Weight of KClO3=Moles of KClO3×Molar mass of KClO3KClO_3 = \text{Moles of } KClO_3 \times \text{Molar mass of } KClO_3 Weight of KClO3=0.043 mol×122.5 g/mol=4.93 g≈1.633 gKClO_3 = \frac{0.04}{3} \text{ mol} \times 122.5 \text{ g/mol} = \frac{4.9}{3} \text{ g} \approx 1.633 \text{ g}

  4. Calculate the weight of KCl produced: From the balanced equation, 2 moles of KClO3KClO_3 produce 2 moles of KClKCl. The mole ratio of KClO3KClO_3 to KClKCl is 2:2 or 1:1. Moles of KCl=Moles of KClO3 reactedKCl = \text{Moles of } KClO_3 \text{ reacted} Moles of KCl=0.043 molKCl = \frac{0.04}{3} \text{ mol} Molar mass of KCl=39+35.5=74.5 g/molKCl = 39 + 35.5 = 74.5 \text{ g/mol} Weight of KCl=Moles of KCl×Molar mass of KClKCl = \text{Moles of } KCl \times \text{Molar mass of } KCl Weight of KCl=0.043 mol×74.5 g/mol=2.983 g≈0.993 gKCl = \frac{0.04}{3} \text{ mol} \times 74.5 \text{ g/mol} = \frac{2.98}{3} \text{ g} \approx 0.993 \text{ g}

Alternatively, by the law of conservation of mass: Weight of KClO3KClO_3 decomposed = Weight of KClKCl produced + Weight of O2O_2 produced Weight of KClKCl produced = Weight of KClO3KClO_3 decomposed - Weight of O2O_2 produced Weight of KClKCl produced ≈1.633 g−0.64 g=0.993 g\approx 1.633 \text{ g} - 0.64 \text{ g} = 0.993 \text{ g}

The results are: (i) Weight of oxygen product = 0.64 g (ii) Weight of KClO3_3 originally taken ≈\approx 1.633 g (iii) Weight of KCl produced ≈\approx 0.993 g