stehiometrija

Submitted by nolenk on Tue, 06/24/2014 - 11:53
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Redukcijom 1,136 grama nekog oksida olova elementarnim ugljikom dobiveno je 0,14 grama ugljen-dioksida.Koji je to oksid?

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milicaradenkovic - Tue, 06/24/2014 - 12:08

PbxOy+C--------------->CO2+Pb

n(CO2)=m/Mr=0.14/44=0.00318 mol

n(O)=2*n(CO2)=2*0.00318=0.00636 mol

m(O)=n(O)*Mr=0.00636*16=0.10176 g

m(Pb)=m(PbxOy)-m(O)=1.136-0.10176=1.03424 g

n(Pb)=m(Pb)/Mr=1.03424/207=0.005 mol

n(Pb):n(O)=0.005:0.00636/:0.005

n(Pb):n(O)=1:1.3/*3

n(Pb):n(O)=3:4 pa je formula oksida Pb3O4