Oxygen gas (O2) can be obtained from the decomposition reaction of potassium chlorate (KClO3): 2KClO3(s) → 2KCl(s) + 3O2(g) a) How many moles of O2 can be produced by the decomposition of 32.8 g KClO3? b) How many grams of KClO3 are required to decompose to produce 50.0 g O2
The Correct Answer and Explanation is:
) How many moles of O2 can be produced by the decomposition of 32.8 g KClO3?
First, we need to use the molar mass of potassium chlorate (KClO3) and the stoichiometry of the reaction to find the number of moles of oxygen gas (O2) produced.
The balanced decomposition reaction is:
2 KClO3(s)→2 KCl(s)+3 O2(g)2 \, \text{KClO}_3(s) \rightarrow 2 \, \text{KCl}(s) + 3 \, \text{O}_2(g)2KClO3(s)→2KCl(s)+3O2(g)
Step 1: Molar Mass of KClO3
- Potassium (K): 39.1 g/mol
- Chlorine (Cl): 35.5 g/mol
- Oxygen (O): 16.0 g/mol
Molar mass of KClO3 = 39.1 + 35.5 + (3 × 16.0) = 122.6 g/mol
Step 2: Convert 32.8 g KClO3 to moles
Moles of KClO3=mass of KClO3molar mass of KClO3=32.8 g122.6 g/mol=0.267 mol KClO3\text{Moles of KClO}_3 = \frac{\text{mass of KClO}_3}{\text{molar mass of KClO}_3} = \frac{32.8 \, \text{g}}{122.6 \, \text{g/mol}} = 0.267 \, \text{mol KClO}_3Moles of KClO3=molar mass of KClO3mass of KClO3=122.6g/mol32.8g=0.267mol KClO3
Step 3: Use Stoichiometry to Find Moles of O2
From the balanced equation, 2 moles of KClO3 produce 3 moles of O2. So, for every 2 moles of KClO3, we get 3 moles of O2. Therefore, we can use the ratio to find the moles of O2:Moles of O2=0.267 mol KClO3×3 mol O22 mol KClO3=0.4005 mol O2\text{Moles of O}_2 = 0.267 \, \text{mol KClO}_3 \times \frac{3 \, \text{mol O}_2}{2 \, \text{mol KClO}_3} = 0.4005 \, \text{mol O}_2Moles of O2=0.267mol KClO3×2mol KClO33mol O2=0.4005mol O2
Thus, 0.4005 moles of O2 can be produced from 32.8 g of KClO3.
b) How many grams of KClO3 are required to decompose to produce 50.0 g O2?
Step 1: Convert Grams of O2 to Moles
First, we calculate the number of moles of oxygen gas (O2) that correspond to 50.0 g:
Molar mass of O2 = 32.0 g/mol.Moles of O2=50.0 g O232.0 g/mol=1.5625 mol O2\text{Moles of O}_2 = \frac{50.0 \, \text{g O}_2}{32.0 \, \text{g/mol}} = 1.5625 \, \text{mol O}_2Moles of O2=32.0g/mol50.0g O2=1.5625mol O2
Step 2: Use Stoichiometry to Find Moles of KClO3
From the balanced reaction, 3 moles of O2 are produced from 2 moles of KClO3. Therefore, we can use the ratio to find the moles of KClO3 required:Moles of KClO3=1.5625 mol O2×2 mol KClO33 mol O2=1.0417 mol KClO3\text{Moles of KClO}_3 = 1.5625 \, \text{mol O}_2 \times \frac{2 \, \text{mol KClO}_3}{3 \, \text{mol O}_2} = 1.0417 \, \text{mol KClO}_3Moles of KClO3=1.5625mol O2×3mol O22mol KClO3=1.0417mol KClO3
Step 3: Convert Moles of KClO3 to Grams
Finally, we convert the moles of KClO3 to grams using the molar mass of KClO3 (122.6 g/mol):Mass of KClO3=1.0417 mol KClO3×122.6 g/mol=127.5 g KClO3\text{Mass of KClO}_3 = 1.0417 \, \text{mol KClO}_3 \times 122.6 \, \text{g/mol} = 127.5 \, \text{g KClO}_3Mass of KClO3=1.0417mol KClO3×122.6g/mol=127.5g KClO3
Thus, 127.5 grams of KClO3 are required to produce 50.0 grams of oxygen gas (O2).
