\(n_{C_2H_2}=\dfrac{8,96}{22,4}=0,4mol\\ a.2C_2H_2+5O_2\xrightarrow[]{t^0}4CO_2+2H_2O\\ n_{CO_2}=2n_{C_2H_2}=0,8mol\\ n_{H_2O}=n_{C_2H_2}=0,4mol\\ V_{CO_2}=0,8.22,4=17,92l\\ m_{H_2O}=0,4.18=7,2g\\ c.n_{O_2}=\dfrac{5}{2}n_{C_2H_2}=1mol\\ V_{KK}=\dfrac{1.22,4}{20\%}\cdot100\%=112l\)
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