Ile moli metanolu przypada na 1 mol metanolanu sodu w roztworze otrzymanym po reakcji 2,3 g sodu z 32 g metanolu.
2CH3OH + 2Na -> 2CH3ONa + H264g 2CH3OH - 46g 2Nax - 2,3g Nax= 3,2g CH3OH
46g 2Na - 2,3g Na108g 2CH3ONa - xx=5,4g
M CH3ONa= 14u+3u+16u+23u= 54u= 54g1mol CH3ONa-----54g n moli-----5,4gn=0,1mola
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2CH3OH + 2Na -> 2CH3ONa + H2
64g 2CH3OH - 46g 2Na
x - 2,3g Na
x= 3,2g CH3OH
46g 2Na - 2,3g Na
108g 2CH3ONa - x
x=5,4g
M CH3ONa= 14u+3u+16u+23u= 54u= 54g
1mol CH3ONa-----54g
n moli-----5,4g
n=0,1mola