A. 11,2.
B. 13,44.
C. 5,60.
D. 8,96.
A
Đáp án A
Ta có: \({{C}_{2}}{{H}_{2\,\,(d\ddot{o})}}={{n}_{{{C}_{2}}A{{g}_{2}}\downarrow }}=\frac{12}{240}=0,05\,(mol);\,\,{{C}_{2}}{{H}_{4}}={{n}_{B{{r}_{2}}}}=\frac{16}{160}=0,1\,(mol)\)
\(\xrightarrow{BTH}2{{n}_{{{C}_{2}}{{H}_{2}}(b\tilde{n})}}+2{{n}_{{{H}_{2}}(b\tilde{n})}}=2n{{C}_{2}}{{H}_{2(d\ddot{o})}}+4{{n}_{{{C}_{2}}{{H}_{4}}}}+2{{n}_{{{H}_{2}}O}}\)
\(\to \underbrace{{{n}_{{{C}_{2}}{{H}_{2}}(b\tilde{n})}}+{{n}_{{{H}_{2}}(b\tilde{n})}}}_{{{n}_{X}}}=\underbrace{{{n}_{{{C}_{2}}{{H}_{2}}(d\ddot{o})}}}_{0,05}+\underbrace{2{{n}_{{{C}_{2}}{{H}_{4}}}}}_{2.0,1}+\underbrace{{{n}_{{{H}_{2}}O}}}_{0,25}=0,5\,(mol)\to {{V}_{X}}=11,2(l)\)
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