Physical Chemistry
Please answer question 8 .plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz
Please answer question 8 .plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz
1 Answer
Arun
As per second equation,
4 moles of KClO3 forms 3 moles of KClO4
Molar mass of KClO3 = 122.5 g
Molar mass of KClO4 = 138.5 g
So, 138.5 x 3 g of KClO4 = 122.5 x 4 g of KClO3
0.395 g of KClO4 = (122.5 x 4 x 0.395)/(138.5 x 3) = 0.466 g of KClO3
So, out of 1.5 g of KClO3, 0.466 g decomposes as per equation 2
1.034 g will decompose as per equation 1
Now, acc to equation 1, 2 moles of KClO3 forms 3 moles of O2
As 1 mole of O2 = 22.4 L at STP
2 x 122.5 g of KClO3 forms 3 x 22.4 L of O2
1.034 g of KClO3 will form (3 x 22.4 x 1.034)/2 x 122.5 = 0.2836 L or 283.6 mL
Correct answer is option A
4 moles of KClO3 forms 3 moles of KClO4
Molar mass of KClO3 = 122.5 g
Molar mass of KClO4 = 138.5 g
So, 138.5 x 3 g of KClO4 = 122.5 x 4 g of KClO3
0.395 g of KClO4 = (122.5 x 4 x 0.395)/(138.5 x 3) = 0.466 g of KClO3
So, out of 1.5 g of KClO3, 0.466 g decomposes as per equation 2
1.034 g will decompose as per equation 1
Now, acc to equation 1, 2 moles of KClO3 forms 3 moles of O2
As 1 mole of O2 = 22.4 L at STP
2 x 122.5 g of KClO3 forms 3 x 22.4 L of O2
1.034 g of KClO3 will form (3 x 22.4 x 1.034)/2 x 122.5 = 0.2836 L or 283.6 mL
Correct answer is option A
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