1.

The half-life for the reaction N_(2)O_(5) to 2 NO_(2) + (1)/(2) O_(2) is 2.4 h at STP . Starting with 10.8 g of N_(2)O_(5) how much oxygen will be obtained after a period of 9.6 h

Answer»

1.5 L
3.36 L
1.05 L
0.07 L

Solution :Moles of `N_(2)O_(5) = (10.8)/(108) = 0.1` and `n = (9.6)/(2.4) = 4`
Here n = NUMBERS of half-life
`N_(2)O_(5) to 2NO_(2) + (1)/(2)O_(2) "" therefore N_(2) = 0.1 xx ((1)/(2))^(n)`
Moles of `N_(2)O_(5)` LEFT `= (0.1)/(16)`
Moles of `N_(2)O_(5)` CHANGED to product = `(0.1 - (0.1)/(16)) = (1.5)/(16)` MOL
Moles of `O_(2)` formed = `(1.5)/(16) xx (1)/(2) = (1.5)/(32)`
Volume of oxygen = `(1.5)/(32) xx 22.4 = 1.05` L.


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