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5,75g Na - 18,5 g CnH₂n⁺₁OH
23g Na = x g CnH₂n⁺₁OH
x= 74 g CnH₂n⁺₁OH
Jest to masa molowa alkoholu:
MCnH₂n⁺₁OH = 12n + 2n + 1 + 16 + 1
74= 14n + 18
56 = 14n
n=4
poszukiwany alkohol to:
C₄H₉OH (butanol)