Дано m(ppanaoh)=100g w(naoh)=20% m(ppafeso4)=200 g w(feso4)=7.6% m(fe(oh) m(naoh)=100*20%/100%=20 g m(feso4)=200*7.6%/100%=15.2 g 20 15.2 x 2naoh+feso4--> fe(oh)2↓+na2so4 152 90 m(naoh)=40 g/mol m(feso4)=152 g/mol m(fe(oh)2=90 g/mol n(naoh)=m/m=20/40 = 0.5 mol n(feso4)=15.2/152=0.1 mol n(naoh)> n(feso4) 15.2/152 = x/90 x=9 g ответ 9 г
alexander4590
13.12.2022
Дано m(al)=540 mg=0.54 g w(прим)=4% hcl n( v( m(чист al)=0.54-(0.54*4%/100%)=0.5184 g 0.5184 x 2al + 6hcl = 2alcl3 + 3h2 2mol 3mol m(al)=27 g/mol , vm=22.4 l/mol n(al)=m/m=0.5184/27=0.0192 mol n(al)=n(alcl3)=0.0192 mol n(alcl3)=0.0192 mol 0.5184/2 = x/3 x=0.7776 л - водорода ответ 0.0192 моль - alcl3 , 0.7776 л - водорода