In the decomposition reaction

1. In the decomposition reaction: 2KClO3(s)——-> 2KCl(s) + 3O2(g), 3.66 g of KClO3 is decomposing.

a. What volume of O2 is collected at STP?

b. What volume of O2 is collected at 30.0 degree C and 745mmHg?

The Correct Answer and Explanation is :

Let’s break down the problem step by step, and then I’ll provide the necessary explanation.

Decomposition Reaction:

The given reaction is:
[ 2 \text{KClO}_3 (s) \rightarrow 2 \text{KCl} (s) + 3 \text{O}_2 (g) ]

Given Data:

• Mass of ( \text{KClO}_3 ) = 3.66 g
• Molar mass of ( \text{KClO}_3 ) = 122.55 g/mol
• Standard Temperature and Pressure (STP): ( 0^\circ C ) and ( 1 \, \text{atm} ) (which equals ( 22.4 \, \text{L/mol} ) for gases)
• Given conditions for second part: 30.0°C and 745 mmHg

Part a: Volume of ( O_2 ) Collected at STP

1. Find moles of ( \text{KClO}_3 ):
   [
   \text{Moles of KClO}_3 = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{3.66 \, \text{g}}{122.55 \, \text{g/mol}} = 0.0298 \, \text{mol}
   ]
2. Use stoichiometry to find moles of ( O_2 ):
   From the balanced equation, 2 moles of ( \text{KClO}_3 ) produce 3 moles of ( O_2 ).
   [
   \text{Moles of O}_2 = \left( \frac{3}{2} \right) \times 0.0298 \, \text{mol} = 0.0447 \, \text{mol}
   ]
3. Use the molar volume at STP to find the volume of ( O_2 ):
   At STP, 1 mole of any ideal gas occupies 22.4 L.
   [
   \text{Volume of O}_2 = 0.0447 \, \text{mol} \times 22.4 \, \text{L/mol} = 1.00 \, \text{L}
   ]

Thus, the volume of ( O_2 ) collected at STP is 1.00 L.

Part b: Volume of ( O_2 ) Collected at 30.0°C and 745 mmHg

1. Use the ideal gas law to calculate the volume at the given conditions:

The ideal gas law equation is:
[
PV = nRT
]
Where:

• ( P ) = pressure in atm = ( \frac{745 \, \text{mmHg}}{760 \, \text{mmHg/atm}} = 0.980 \, \text{atm} )
• ( V ) = volume of ( O_2 ) (to be determined)
• ( n ) = moles of ( O_2 ) = 0.0447 mol
• ( R ) = ideal gas constant = 0.0821 L·atm/mol·K
• ( T ) = temperature in Kelvin = ( 30.0 + 273.15 = 303.15 \, \text{K} )

1. Rearrange the ideal gas law to solve for ( V ):
   [
   V = \frac{nRT}{P}
   ]
   Substituting in the values:
   [
   V = \frac{(0.0447 \, \text{mol}) (0.0821 \, \text{L·atm/mol·K}) (303.15 \, \text{K})}{0.980 \, \text{atm}} = 1.16 \, \text{L}
   ]

Thus, the volume of ( O_2 ) collected at 30.0°C and 745 mmHg is 1.16 L.

Summary:

• a) The volume of ( O_2 ) collected at STP is 1.00 L.
• b) The volume of ( O_2 ) collected at 30.0°C and 745 mmHg is 1.16 L.

Now, I’ll generate a visual representation of the key data and the calculations.

Here is the visual representation of the volume of ( O_2 ) collected under the two different conditions: STP and 30°C at 745 mmHg. As you can see, the volume increases slightly when the temperature rises and the pressure is reduced.