1.

Consider the following reaction : 8H_(2) (g) + S_(8) (l) rarr 8H_(2) S (g) is carried out at 127^(@) C and 20 atm. Then what mass of S_(8) would be required to produce 8L of H_(2) S (g) under these conditions. [Take : R = 0.08 " atm " L//"mole" - K)

Answer»

1600 gm
1280 gm
20 gm
160 gm

Solution :`8 H_(2) (g) + S_(8) (l) rarr 8H_(2) S (g)`
`T = 400 k`
`P = 20` atm
`V = 8L`
Using `PV = nRT`
`n = 5` mole
For PRODUCING 5 moles of `H_(2)S`,
Moles of `S_(8)` required `= (1)/(8) xx 5 = (5)/(8)` moles
Mass of `S_(8)` required `= (5)/(8) xx (32 xx 8) = 160` gm


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