treba mi pomoc oko sledeceg zadatka:Kolika je zapremina 26% sumporne kiseline (ro je 1,19 g/cm kubnom) potrebna za rastvaranje 50 grama gvozđa?
Fe+H2SO4--------------->FeSO4+H2
m(Fe)=50 g
n(Fe)=m(Fe)/Mr=50/55.85=0.895 mol
n(Fe):n(H2SO4)=1:1
n(H2SO4)=n(Fe)=0.895 mol
m(H2SO4)=n(H2SO4)*Mr=0.895*98=87.71 g
mr=m(H2SO4)/w(H2SO4)=87.71/0.26=337.35 g
V(H2SO4)=mr/ro=337.35/1.19=283.48 cm3 26% H2SO4
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gvozdje
Fe+H2SO4--------------->FeSO4+H2
m(Fe)=50 g
n(Fe)=m(Fe)/Mr=50/55.85=0.895 mol
n(Fe):n(H2SO4)=1:1
n(H2SO4)=n(Fe)=0.895 mol
m(H2SO4)=n(H2SO4)*Mr=0.895*98=87.71 g
mr=m(H2SO4)/w(H2SO4)=87.71/0.26=337.35 g
V(H2SO4)=mr/ro=337.35/1.19=283.48 cm3 26% H2SO4