m(ppa)=78,3+47,4=125,7г
m(FeCl3)=78,3×0,15=11,745г
ω(FeCl3)=11,745:125,7=0,0934×100%=9,34%
m(NaOH)=47,4×0,09=4,266г
ω(NaOH)=4,266:125,7=0,034×100%=3,4%
m(H2O)=125,7-11,645-4,266=109,789г
ω(H2O)=109,789:125,7;=0,8734×100%=87,34%
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Answers & Comments
m(ppa)=78,3+47,4=125,7г
m(FeCl3)=78,3×0,15=11,745г
ω(FeCl3)=11,745:125,7=0,0934×100%=9,34%
m(NaOH)=47,4×0,09=4,266г
ω(NaOH)=4,266:125,7=0,034×100%=3,4%
m(H2O)=125,7-11,645-4,266=109,789г
ω(H2O)=109,789:125,7;=0,8734×100%=87,34%