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Sagot :
Certainly! Let's solve this step-by-step.
### Step 1: Calculate the moles of KClO_3
First, we need to find the molar mass of KClO_3. The molar masses of the elements are approximately:
- Potassium (K): 39.10 g/mol
- Chlorine (Cl): 35.45 g/mol
- Oxygen (O): 16.00 g/mol (since there are 3 oxygen atoms, we multiply by 3)
So, the molar mass of potassium chlorate (KClO_3) is:
[tex]\[ 39.10 + 35.45 + (16.00 \times 3) = 122.55 \, \text{g/mol} \][/tex]
Given that the mass of KClO_3 is 4.35 grams, we can calculate the number of moles:
[tex]\[ \text{moles of } KClO_3 = \frac{\text{mass of } KClO_3}{\text{molar mass of } KClO_3} = \frac{4.35 \, \text{g}}{122.55 \, \text{g/mol}} = 0.0355 \, \text{mol} \][/tex]
### Step 2: Determine the moles of O_2 produced
From the balanced chemical equation:
[tex]\[ 2 \, \text{KClO}_3 \rightarrow 2 \, \text{KCl} + 3 \, \text{O}_2 \][/tex]
The molar ratio of KClO_3 to O_2 is 2:3. Therefore, for every 2 moles of KClO_3 decomposed, 3 moles of O_2 are produced. We can use this ratio to find the moles of oxygen gas:
[tex]\[ \text{moles of } O_2 = \left(\frac{3}{2}\right) \times \text{moles of } KClO_3 = \left(\frac{3}{2}\right) \times 0.0355 \, \text{mol} = 0.0532 \, \text{mol} \][/tex]
### Step 3: Calculate the volume of O_2 gas using the Ideal Gas Law
The Ideal Gas Law equation is given by:
[tex]\[ PV = nRT \][/tex]
where:
- [tex]\( P \)[/tex] is the pressure (0.75 atm)
- [tex]\( V \)[/tex] is the volume (which we need to find)
- [tex]\( n \)[/tex] is the number of moles of O_2 (0.0532 mol)
- [tex]\( R \)[/tex] is the ideal gas constant (0.0821 L·atm/(mol·K))
- [tex]\( T \)[/tex] is the temperature in Kelvin
First, convert the temperature from Celsius to Kelvin:
[tex]\[ T = 27^\circ C + 273.15 = 300.15 \, K \][/tex]
Now, rearrange the Ideal Gas Law to solve for [tex]\( V \)[/tex]:
[tex]\[ V = \frac{nRT}{P} = \frac{(0.0532 \, \text{mol}) \times (0.0821 \, \frac{\text{L·atm}}{\text{mol·K}}) \times (300.15 \, \text{K})}{0.75 \, \text{atm}} = 1.749 \, \text{L} \][/tex]
### Final Answer:
The volume of the O_2 gas produced is approximately [tex]\( 1.75 \, \text{L} \)[/tex].
Thus, a 4.35 gram sample of KClO_3, when decomposed, produces oxygen gas that occupies a volume of 1.75 liters under a pressure of 0.75 atm and a temperature of [tex]\(27^\circ C\)[/tex].
### Step 1: Calculate the moles of KClO_3
First, we need to find the molar mass of KClO_3. The molar masses of the elements are approximately:
- Potassium (K): 39.10 g/mol
- Chlorine (Cl): 35.45 g/mol
- Oxygen (O): 16.00 g/mol (since there are 3 oxygen atoms, we multiply by 3)
So, the molar mass of potassium chlorate (KClO_3) is:
[tex]\[ 39.10 + 35.45 + (16.00 \times 3) = 122.55 \, \text{g/mol} \][/tex]
Given that the mass of KClO_3 is 4.35 grams, we can calculate the number of moles:
[tex]\[ \text{moles of } KClO_3 = \frac{\text{mass of } KClO_3}{\text{molar mass of } KClO_3} = \frac{4.35 \, \text{g}}{122.55 \, \text{g/mol}} = 0.0355 \, \text{mol} \][/tex]
### Step 2: Determine the moles of O_2 produced
From the balanced chemical equation:
[tex]\[ 2 \, \text{KClO}_3 \rightarrow 2 \, \text{KCl} + 3 \, \text{O}_2 \][/tex]
The molar ratio of KClO_3 to O_2 is 2:3. Therefore, for every 2 moles of KClO_3 decomposed, 3 moles of O_2 are produced. We can use this ratio to find the moles of oxygen gas:
[tex]\[ \text{moles of } O_2 = \left(\frac{3}{2}\right) \times \text{moles of } KClO_3 = \left(\frac{3}{2}\right) \times 0.0355 \, \text{mol} = 0.0532 \, \text{mol} \][/tex]
### Step 3: Calculate the volume of O_2 gas using the Ideal Gas Law
The Ideal Gas Law equation is given by:
[tex]\[ PV = nRT \][/tex]
where:
- [tex]\( P \)[/tex] is the pressure (0.75 atm)
- [tex]\( V \)[/tex] is the volume (which we need to find)
- [tex]\( n \)[/tex] is the number of moles of O_2 (0.0532 mol)
- [tex]\( R \)[/tex] is the ideal gas constant (0.0821 L·atm/(mol·K))
- [tex]\( T \)[/tex] is the temperature in Kelvin
First, convert the temperature from Celsius to Kelvin:
[tex]\[ T = 27^\circ C + 273.15 = 300.15 \, K \][/tex]
Now, rearrange the Ideal Gas Law to solve for [tex]\( V \)[/tex]:
[tex]\[ V = \frac{nRT}{P} = \frac{(0.0532 \, \text{mol}) \times (0.0821 \, \frac{\text{L·atm}}{\text{mol·K}}) \times (300.15 \, \text{K})}{0.75 \, \text{atm}} = 1.749 \, \text{L} \][/tex]
### Final Answer:
The volume of the O_2 gas produced is approximately [tex]\( 1.75 \, \text{L} \)[/tex].
Thus, a 4.35 gram sample of KClO_3, when decomposed, produces oxygen gas that occupies a volume of 1.75 liters under a pressure of 0.75 atm and a temperature of [tex]\(27^\circ C\)[/tex].
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