What weight of CO is required to form `Re_(2)(CO)_(10)` from 2.50 g fo f`Re_(2)O_(7)` according to the unbalanced reaction:
`Re_(2)O_(7) + CO rarr Re_(2)(CO)_(10)+CO_(2)`
(Re =186 ,C =12 and O=16).Give your answer to the nearest integar.

Explanation

Suppose the relative moles of each reactant and product are as follows (just for convenience)
`{:(Re_(2)O_(7)" " + CO " " to " "Re_(2) (CO)_(10)" + "CO_(2)),(" a moles b moles c moles d moles "):}`
Applying POAC for Re atoms,
`{:(" moles of Re in " Re_(2) O_(7) =" moles of Re in " RE_(2) (CO)_(10)),(" "2 xx "moles of " Re_(2) O_(7) = 2 xx "moles of " Re_(2) (CO)_(10)),(" Ra = 2c"),(" or a = c" ):}` . . . (i)
Applying POAC for C atoms,
moles of C atoms in CO = moles of C in `Re_(2) (CO)_(10) `
+ moles of C in ` CO_(2)`
`xx` moles of CO ` = 10 xx ` moles of `Re_(2) (CO)_(10) + 1 xx` moles of `CO_(2)`
or b = 10 c + d . . . (ii)
moles of O in `Re_(2) O_(7) ` + moles of O in CO
= moles of O in `Re_(2) (CO)_(10) + ` moles of O in `CO_(2)`
`7 xx ` moles of `Re_(2)O_(7) + 1 xx ` moles of CO
`= 10 xx` moles of `Re_(2) (CO)_(10) + 2 xx ` moles of `CO_(2)`
or `7a + b = 10 c + 2 d" "` . . . (iii)
From the eqns . (i),(ii)and (iii) , we get 17 a = b
i.e., `17 xx ` moles of `Re_(2) O_(7)` = moles of CO
`17 xx (2 . 50)/( 484,.4) = ("wt . of CO in g")/(28) [{:(mol. wt. of Re_(2) O_(7) = 484.4),(" mol. wt. of CO = 28"):}] `
Wt . of CO = 2 . 46 g