Answer:
1.04L
Explanation:
First, we'll begin by calculating the number of mole of KClO3 present in 5.45g of the compound.
This is illustrated below:
Molar Mass of KClO3 = 39 + 35.5 + (16x3) = 39 + 35.5 + 48 = 122.5g/mol
Mass of KClO3 from the question = 5.45g
Mole of KClO3 =?
Number of mole = Mass /Molar Mass
Mole of KClO3 = 5.45/122.5
Mole of KClO3 = 0.044mole
The equation for the reaction is given below:
2KClO3 → 2KCl + 3O2
From the equation above,
2 moles of KClO3 produced 3 moles of O2.
Therefore, 0.044 mole of KClO3 will produce = (0.044 x 3)/2 = 0.066 mole of O2.
Now, we can obtain the volume of O2 given off by using the ideal gas equation as illustrated below:
Data obtained from the question include:
P (pressure) = 1.58 atm
T (temperature) = 32°C = 32 + 273 = 305K
V (volume) =?
n (number of mole of O2) = 0.066 mole
R (gas constant) = 0.082atm.L/Kmol
Using the ideal gas equation PV = nRT, the volume of O2 given off can be obtained as follow:
PV = nRT
1.58 x V = 0.066 x 0.082 x 305
Divide both side by 1.58
V = (0.066 x 0.082 x 305)/1.58
V = 1.04L
Therefore, the volume of O2 given off is 1.04L
