Asked by G.C.
The problem asked to find the amount (in moles) of reactant left.
.223moles FeS and .652moles of HCl
FeS+2HCl--> FeCl2+H2S
.223molFeS(1moleFeCl2/1moleFeS)=.223moleFeCl2
.652molHCl(1moleFeCl2/2molHCl)=.326moleFeCl2
.223/.326x100=68.4% used HCL
100-68.4=31.6% remaining
31.6%x(.652molHCl)=0.206mol HCl remaning
I was just wondering if I did it correctly.
.223moles FeS and .652moles of HCl
FeS+2HCl--> FeCl2+H2S
.223molFeS(1moleFeCl2/1moleFeS)=.223moleFeCl2
.652molHCl(1moleFeCl2/2molHCl)=.326moleFeCl2
.223/.326x100=68.4% used HCL
100-68.4=31.6% remaining
31.6%x(.652molHCl)=0.206mol HCl remaning
I was just wondering if I did it correctly.
Answers
Answered by
bobpursley
yes.
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