The work done then 6.5 g of zinc reacts with dil.HC1 is an open beaker at 298 K is

Zn+2HCl⟶ZnCl2+H2(g)                     …(1)           1mol                  1mol      Since H2 gas drives back the atmosphere in an open beaker, work is done by the gas which is negative.            ∴ W=−Pext ×ΔV Since initial volume zero  so  ΔV=  Final volume          initial volume;∆ V = Final volume But           ΔV=nRTPext       W=−Pext ×nRTPext =−nRTn= no. of mol of Zn= wt. of Zn at. wt. of Zn=6.565=110From equation (1),           1 mol of Zn≡1 mol of H2;110 mol  of                Zn≡110mol of H2W=−nRT;W=−110mol×8.314JK−mol×298K=-247.76J