giúp mih 2 bài này vs ạ đag cần gấp!!!!!!

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giúp mih 2 bài này vs ạ đag cần gấp!!!!!!

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    Câu 6:

    `a)`

    `C_2H_4 + Br_2 -> C_2H_4Br_2`

    $CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O$
    $C_2H_4 + 3O_2 \xrightarrow{t^o} 2CO_2 + 2H_2O$
    `CO_2 + Ca(OH)_2 -> CaCO_3 + H_2O`

    `b)`

    `n_{Br_2} = (32)/(160) = 0,2(mol)`

    `n_{CaCO_3} = (50)/(100) = 0,5(mol)`

    Theo PT: 

    `n_{C_2H_4} = n_{Br_2} = 0,2(mol)`

    `n_{CO_2} = n_{CaCO_3} = 0,5(mol)`

    `n_{CO_2} = 2n_{C_2H_4} + n_{CH_4}`

    `=> n_{CH_4} = 0,5 – 0,2.2 = 0,1(mol)`

    `\%V_{CH_4} = (0,1)/(0,1 + 0,2) .100\% = 33,33\%`

    `\%V_{C_2H_4} = 100\% – 33,33\% = 66,67\%`

    Câu 7:

    `a)`

    `n_{hh} = (8,96)/(22,4) = 0,4(mol)`

    `C_nH_{2n + Br_2 -> C_nH_{2n}Br_2`

    `n_{Br_2} = (48)/(160) = 0,3(mol)`

    Theo PT: `n_{C_nH_{2n}} = n_{Br_2} = 0,3(mol)`

    `=> n_{CH_4} =0,4 – 0,3 = 0,1(mol)`

    `\%V_{CH_4} = (0,1)/(0,4) .100\% = 25\%`

    `\%V_{C_nH_{2n}} = 100\% – 25\% = 75\%`

    `b)`

    `n_{CO_2} = (30,8)/(44) = 0,7(mol)`

    $CH_4 + O_2 \xrightarrow{t^o} CO_2 + 2H_2O$
    $2C_nH_{2n} + 3nO_2 \xrightarrow{t^o} 2nCO_2 + 2nH_2O$
    Theo PT: `n_{CO_2} = n_{CH_4} + n.n_{C_nH_{2n}}`

    `=> 0,1 + 0,2n = 0,7`

    `<=> n = 3`

    `=> X: C_3H_6`

    CTCT: `CH_3-CH=CH_2`

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