a, PT: \(2C_2H_5OH+2K\rightarrow2C_2H_5OK+H_2\)
b, Ta có: VC2H5OH = 56,25.46% = 25,875 (ml)
⇒ mC2H5OH = 25,875.0,8 = 20,7 (g)
c, Có: \(n_{C_2H_5OH}=\dfrac{20,7}{46}=0,45\left(mol\right)\)
Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{C_2H_5OH}=0,225\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,225.22,4=5,04\left(l\right)\)
Bạn tham khảo nhé!