\(V_{C_2H_5OH}=\dfrac{36,8.90}{100}=33,12ml\\ m_{C_2H_5OH}=33,12.0,8=26,496g\\ n_{C_2H_5OH}=\dfrac{26,496}{46}=0,576mol\\ 2C_2H_5OH+2K\rightarrow2C_2H_5OK+H_2\\ n_{H_2}=\dfrac{1}{2}n_{C_2H_5OH}=0,288mol\\ V_{H_2}=0,228.22,4=6,4512l\)
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