Ustal wzór alkanolu wiedząc, że z 11g tego związku w reakcji z sodem otrzymano 1,4dm3 wodoru (warunki normalne)
CnH2n+1OH + Na---->CnH2n+1ONa + ½H2
11g alkanolu-------1,4dm3 H2
xg alkanolu--------11,2dm3 H2
x = 88g alkanolu
CnH2n+1OH=88u
12n+2n+1+16+1 = 88
14n = 70
n = 5
Szukany alkanol: C5H11OH (pentanol)
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CnH2n+1OH + Na---->CnH2n+1ONa + ½H2
11g alkanolu-------1,4dm3 H2
xg alkanolu--------11,2dm3 H2
x = 88g alkanolu
CnH2n+1OH=88u
12n+2n+1+16+1 = 88
14n = 70
n = 5
Szukany alkanol: C5H11OH (pentanol)