When phosphours `(P_(4))` is heated in limited amount of `O_(2).P_(4)O_(6)` (tetraphosphorous hexaoxide) is obtained, and in excess of `O_(2)`, `P_(4) O_(10)` (tetraphosphours decaoxide) is obtained.
i. `P + 3 O_(2) rarr P_(4) O_(6)`, ii. `P_(4) + 5O_(2) rarr P_(4)O_(10)`
how many moles of O_(2) left unreached intially in reaction (i)when 2g of phosphours reacts with 2g of 'O2?

Options

A. 0.0145 mol
B. 0.0072 mol
C. 0.029
D. 0.0048

Answer

B

Explanation

Mol of `O_(2) = (2.0)/(31 xx 4) = 0.016 "mol"`
This is not enough `O_(2)` to produce `P_(4) O_(10)`, which would have required `5 xx 0.016 = 0.08` mol `O_(2)`. Thus,
`(0.016 "mol" P_(4)) ((3 "mol"O_(2))/("mol"P_(4)))`
`= 0.048 "mol" O_(2)` used to make `P_(4)O_(6)`
Mol of `O_(2)` unreacted `= (0.0625 - 0.048) = 0.0145 "mol" O_(2)`
This `O_(2)` can react according to the equation
`P_(4) O_(6) + 2 O_(2) rarr P_(4) O_(10`
Then,
`(0.0145 "mol" O_(2)) ((1 "mol" P_(4) O_(10))/(2 "mol" O_(2))) = 0.00721 "mol" P_(4) O_(10)`
the `0.016 "mol of" P_(4) O_(6)` orginally formed, `0.00725`
mol is converted to `P_(4) O_(10)`, leaving `(0.016 - 0.00721) = 0.009 "mol" P_(4) O_(6)`
`(Mw "of" P_(4) O_(6) = 220)`
Weight of `P_(4) O_(6) = 0.009 xx 220 = 1.98 g P_(4) O_(6)`