When heated KCLO3 decomposes into KCI and O2, but if this reaction produced 36.2g of KCL, how much O2 was produced (in grams)?
The Correct Answer and Explanation is:
Given reaction: 2KClO3→2KCl+3O22 \text{KClO}_3 \rightarrow 2 \text{KCl} + 3 \text{O}_2
Step 1: Calculate moles of KCl produced
- Molar mass of KCl:
- K = 39.1 g/mol
- Cl = 35.45 g/mol
- Given mass of KCl = 36.2 g
moles of KCl=36.2 g74.55 g/mol=0.4855 mol\text{moles of KCl} = \frac{36.2 \text{ g}}{74.55 \text{ g/mol}} = 0.4855 \text{ mol}
Step 2: Use stoichiometry to find moles of O₂ produced
From the balanced equation: 2 mol KClO3→2 mol KCl+3 mol O22 \text{ mol KClO}_3 \rightarrow 2 \text{ mol KCl} + 3 \text{ mol O}_2
Mole ratio: 3 mol O22 mol KCl=1.5 mol O2 per mol KCl\frac{3 \text{ mol O}_2}{2 \text{ mol KCl}} = 1.5 \text{ mol O}_2 \text{ per mol KCl}
So, moles of O2=0.4855×1.5=0.72825 mol\text{moles of } O_2 = 0.4855 \times 1.5 = 0.72825 \text{ mol}
Step 3: Calculate mass of O₂ produced
- Molar mass of O2=2×16.00=32.00 g/molO_2 = 2 \times 16.00 = 32.00 \, \text{g/mol}
mass of O2=0.72825×32.00=23.304 g\text{mass of } O_2 = 0.72825 \times 32.00 = 23.304 \text{ g}
Final answer:
23.3 g of O2 produced\boxed{23.3 \text{ g of } O_2 \text{ produced}}
Explanation
When potassium chlorate (KClO3\text{KClO}_3) is heated, it undergoes a decomposition reaction forming potassium chloride (KCl) and oxygen gas (O2\text{O}_2). The balanced chemical equation is: 2KClO3→2KCl+3O22 \text{KClO}_3 \rightarrow 2 \text{KCl} + 3 \text{O}_2
This tells us that 2 moles of potassium chlorate produce 2 moles of potassium chloride and 3 moles of oxygen gas. The molar relationship between KCl and O₂ is therefore 2:3.
Given that 36.2 grams of KCl are formed, the first step is to convert this mass to moles by dividing by the molar mass of KCl (approximately 74.55 g/mol). This gives us about 0.4855 moles of KCl.
Using the stoichiometric ratio from the balanced equation, for every 2 moles of KCl, 3 moles of O₂ are produced. Therefore, the moles of O₂ produced is: 0.4855 mol KCl×3 mol O22 mol KCl=0.72825 mol O20.4855 \, \text{mol KCl} \times \frac{3 \, \text{mol O}_2}{2 \, \text{mol KCl}} = 0.72825 \, \text{mol O}_2
Finally, to find the mass of oxygen gas produced, multiply the moles of oxygen by its molar mass (32.00 g/mol): 0.72825×32.00=23.3 g0.72825 \times 32.00 = 23.3 \text{ g}
This calculation assumes the reaction goes to completion with no side reactions or losses. Therefore, the oxygen produced from the decomposition of potassium chlorate that yields 36.2 g of potassium chloride is approximately 23.3 grams.
