2.Koliko je g olovo (II)-hlorida dobijeno u reakciji 5 g Pb(II)-nitrata sa NaCl ako je prinos 63%?
Pb(NO3)2+2NaCl------------>PbCl2+2NaNO3
m(olovo(II)nitrata)=5 g
n(olovo(II)nitrata)=m(olovo(II)nitrata)/Mr=5/331=0.015 mol
n(PbCl2)=n(Pb(NO3)2)=0.015 mol
m(PbCl2)=n(PbCl2)*Mr=0.015*278=4.17 a posto je prinos 63% onda je masa jednaka
m1(PbCl2)=m(PbCl2)*w=4.17*063=2.63 g
hvala
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Pb(NO3)2+2NaCl------------>Pb
Pb(NO3)2+2NaCl------------>PbCl2+2NaNO3
m(olovo(II)nitrata)=5 g
n(olovo(II)nitrata)=m(olovo(II)nitrata)/Mr=5/331=0.015 mol
n(PbCl2)=n(Pb(NO3)2)=0.015 mol
m(PbCl2)=n(PbCl2)*Mr=0.015*278=4.17 a posto je prinos 63% onda je masa jednaka
m1(PbCl2)=m(PbCl2)*w=4.17*063=2.63 g
hvala
hvala