The work done in an open vessel when 112gm of fe reacts with dilute hcl at 300 k
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The reaction involved is:
Fe +2HCl ---------->FeCl2 +H2
Work done = P(V2-V1)
Since mole of Fe used =112/56 =2
The moles of H2 formed =2
P*(VH2-Vi) = P*VH2
As Vi =0
For H2 P* VH2 = nRT
VH2 = nRT/P
W =P*nRT/P = nRT = 2*2*300=1200 cal or 1.2 kcal
Fe +2HCl ---------->FeCl2 +H2
Work done = P(V2-V1)
Since mole of Fe used =112/56 =2
The moles of H2 formed =2
P*(VH2-Vi) = P*VH2
As Vi =0
For H2 P* VH2 = nRT
VH2 = nRT/P
W =P*nRT/P = nRT = 2*2*300=1200 cal or 1.2 kcal
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