Decomposition reaction of KClO3 can be used to produce an oxygen gas. (The other reaction product is solid KCl). a. Write a balanced chemical equation for this reaction. b. What mass of KClO3 must be decomposed to produce 126 L of oxygen gas at 133oC and 0.880 atm?
The Correct Answer and Explanation is:
To solve this problem, we will go step by step for both parts.
a. Balanced Chemical Equation:
The decomposition reaction of potassium chlorate (KClO₃) to produce potassium chloride (KCl) and oxygen gas (O₂) is represented by the following balanced chemical equation:2KClO3(s)→2KCl(s)+3O2(g)2KClO₃(s) \rightarrow 2KCl(s) + 3O₂(g)2KClO3(s)→2KCl(s)+3O2(g)
This shows that 2 moles of KClO₃ decompose to form 2 moles of KCl and 3 moles of oxygen gas.
b. Mass of KClO₃ required to produce 126 L of oxygen gas:
We can use the Ideal Gas Law equation to calculate the mass of KClO₃ required.
The Ideal Gas Law is:PV=nRTPV = nRTPV=nRT
Where:
- P = pressure in atm
- V = volume in liters (L)
- n = number of moles of gas
- R = ideal gas constant (0.0821 L·atm / mol·K)
- T = temperature in Kelvin (K)
Step 1: Convert the given temperature to Kelvin.
Given temperature = 133°CT(K)=133+273=406 KT(K) = 133 + 273 = 406 \text{ K}T(K)=133+273=406 K
Step 2: Rearrange the Ideal Gas Law to find the number of moles of O₂.
n=PVRTn = \frac{PV}{RT}n=RTPV
Substitute the given values:
- P = 0.880 atm
- V = 126 L
- R = 0.0821 L·atm / mol·K
- T = 406 K
n=(0.880)(126)(0.0821)(406)=110.8833.357≈3.32 mol O₂n = \frac{(0.880)(126)}{(0.0821)(406)} = \frac{110.88}{33.357} \approx 3.32 \text{ mol O₂}n=(0.0821)(406)(0.880)(126)=33.357110.88≈3.32 mol O₂
Step 3: Use stoichiometry to relate moles of O₂ to moles of KClO₃.
From the balanced chemical equation, we know that 3 moles of O₂ are produced by 2 moles of KClO₃. Therefore:moles of KClO₃=23×3.32=2.21 mol KClO₃\text{moles of KClO₃} = \frac{2}{3} \times 3.32 = 2.21 \text{ mol KClO₃}moles of KClO₃=32×3.32=2.21 mol KClO₃
Step 4: Convert moles of KClO₃ to grams.
To convert moles of KClO₃ to grams, use the molar mass of KClO₃. The molar mass of KClO₃ is:M(KClO3)=39.1+35.5+(3×16)=122.6 g/molM(KClO₃) = 39.1 + 35.5 + (3 \times 16) = 122.6 \text{ g/mol}M(KClO3)=39.1+35.5+(3×16)=122.6 g/mol
Now, calculate the mass of KClO₃:mass of KClO₃=2.21 mol×122.6 g/mol≈270.5 g\text{mass of KClO₃} = 2.21 \text{ mol} \times 122.6 \text{ g/mol} \approx 270.5 \text{ g}mass of KClO₃=2.21 mol×122.6 g/mol≈270.5 g
Final Answer:
The mass of KClO₃ required to produce 126 L of oxygen gas at 133°C and 0.880 atm is approximately 270.5 g.
