can jup day

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namnguyen_94

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Ta có: Gọi CT là [TEX]C_nH2n ==> nCO_2 = nH_2O = 0,1.n mol[/TEX]
==> m(dd sau) = 100 + 6,2.n gam
+ [TEX]nNaOH = 0,5405 mol ==> nNaOH ( sau ) = 0,5405 - 0,2.n[/TEX]
==> [tex]\frac{40.(0,5405 - 0,2.n)}{100 + 6,2.n} = 0,05[/tex]
==> [tex]n = 2 ==> C_2H_4[/tex]
 
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