koliko grama natrijum metoksida je potrebno da bi se u reakciji sa odgovarajucim alkil halogenidom nagradilo 4,6g dimetil-etra?
Na-O-CH3 + CH3Cl --> CH3-O-CH3 (dimetil-etar) + NaCl
m(CH3-O-CH3)=4,6g M(CH3-O-CH3)=46g/mol
m(NaOCH3)=? M(NaOCH3)=54g/mol
n(CH3-O-CH3):n(NaOCH3)=1:1
m(NaOCH3)=(4,6*54)/46=5,4g
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Na-O-CH3 + CH3Cl -->
Na-O-CH3 + CH3Cl --> CH3-O-CH3 (dimetil-etar) + NaCl
m(CH3-O-CH3)=4,6g
M(CH3-O-CH3)=46g/mol
m(NaOCH3)=?
M(NaOCH3)=54g/mol
n(CH3-O-CH3):n(NaOCH3)=1:1
m(NaOCH3)=(4,6*54)/46=5,4g