\(n_{H_2}=\dfrac{0,61975}{24,79}=0,025\left(mol\right)\)
PTHH: \(2C_2H_5OH+2Na\rightarrow2C_2H_5ONa+H_2\)
0,05<----------------------------0,025
=> m = 0,05.46 = 2,3 (g)
=> D
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