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giải đáp
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Toán Ankan(2).
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$C_nH_{2n+2}+\frac{3n+1}{2}O_2\rightarrow nCO_2+(n+1)H_2O$
$n_{O_2}=0,1625 mol\Rightarrow 0,1625=\frac{1,45}{14n+2}.\frac{3n+1}{2}\Rightarrow n=4\Rightarrow C_4H_{10}$
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sửa đổi
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Toán Ankan(4).
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$C_nH_{2n+2}+\frac{3n+1}{2}O_2\rightarrow nCO_2+(n+1)H_2O$ $a na$$\Rightarrow \begin{cases}a=\frac{7,2}{14n+2} \\ na=0,5 \end{cases}\Rightarrow n=5\Rightarrow C_5H_{10}\Rightarrow CH_3-CH_2-CH(CH_3)-CH_3$
$C_nH_{2n+2}+\frac{3n+1}{2}O_2\rightarrow nCO_2+(n+1)H_2O$ $a na$$\Rightarrow \begin{cases}a=\frac{7,2}{14n+2} \\ na=0,5 \end{cases}\Rightarrow n=5\Rightarrow C_5H_{12}\Rightarrow CH_3-CH_2-CH(CH_3)-CH_3$
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giải đáp
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Toán Ankan(3).
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$C_nH_{2n+2}+\frac{3n+1}{2}O_2\rightarrow nCO_2+(n+1)H_2O$
$\Rightarrow \frac{1,8}{14n+2}.[44n-18(n+1)]=2,8\Rightarrow n=5\Rightarrow C_5H_{12}$
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giải đáp
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Toán Ankan(4).
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$C_nH_{2n+2}+\frac{3n+1}{2}O_2\rightarrow nCO_2+(n+1)H_2O$
$a na$
$\Rightarrow \begin{cases}a=\frac{7,2}{14n+2} \\ na=0,5 \end{cases}\Rightarrow n=5\Rightarrow C_5H_{12}\Rightarrow CH_3-CH_2-CH(CH_3)-CH_3$
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giải đáp
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Toán Ankan(5).
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$C_nH_{2n+2}+\frac{3n+1}{2}O_2\rightarrow nCO_2+(n+1)H_2O$
$\Rightarrow 1+\frac{3n+1}{2}=2.n\Rightarrow n=2\Rightarrow C_2H_6$
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giải đáp
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Toán Ankan(6).
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$C_nH_{2n+2}+\frac{3n+1}{2}O_2\rightarrow nCO_2+(n+1)H_2O$
$\Rightarrow 1+\frac{3n+1}{2}=n+(n+1)\Rightarrow n=1\Rightarrow CH_4$
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giải đáp
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Toán Ankan(7).
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$C_nH_{2n+2}+\frac{3n+1}{2}O_2\rightarrow nCO_2+(n+1)H_2O$
$\Rightarrow 1+\frac{3n+1}{2}=2.n\Rightarrow n=2\Rightarrow C_2H_6$
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giải đáp
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Toán Ankan(8).
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$C_nH_{2n+2}\rightarrow C_nH_{2n}+H_2$
$M_Y=44.0,5=22\Rightarrow M_X=2.22=44\Rightarrow C_3H_6$
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sửa đổi
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Toán Ankan(9).
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Giả sử ban đầu có $1 mol C_4H_{10}\Rightarrow n_{hh}=\frac{58}{36,25}\Rightarrow n_{C_4H_{10 dư}}=1-(\frac{58}{36,25} -1)=0,4 mol$$\Rightarrow H=40\%$
Giả sử ban đầu có $1 mol C_4H_{10}\Rightarrow n_{hh}=\frac{58}{36,25}\Rightarrow n_{C_4H_{10 pư}}=(\frac{58}{36,25} -1)=0,6 mol$$\Rightarrow H=60\%$
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giải đáp
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Toán Ankan(9).
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Giả sử ban đầu có $1 mol C_4H_{10}\Rightarrow n_{hh}=\frac{58}{36,25}\Rightarrow n_{C_4H_{10 pư}}=(\frac{58}{36,25} -1)=0,6 mol$
$\Rightarrow H=60\%$
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giải đáp
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Toán Ankan(10).
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$\begin{cases}4.560=4.V_{C_4H_{10}}+2.V_{C_2H_6}+2.V_{C_2H_4}+V_{CH_4} +3.V_{C_3H_6}+4.V_{C_4H_8} (1)\\ 10.560=10.V_{C_4H_{10}}+6.V_{C_2H_6}+4.V_{C_2H_4}+4.V_{CH_4} +6.V_{C_3H_6}+8.V_{C_4H_8}+2.V_{H_2} (2)\\1010=V_{C_4H_{10}}+V_{C_2H_6}+V_{C_2H_4}+V_{CH_4} +V_{C_3H_6}+V_{C_4H_8}+V_{H_2} (3) \end{cases}$
$\Rightarrow (2)-2.(3)\Rightarrow 3580=8.V_{C_4H_{10}}+4.V_{C_2H_6}+2.V_{C_2H_4}+2.V_{CH_4} +4.V_{C_3H_6}+6.V_{C_4H_8} (4)$
$\Rightarrow (1)-\frac{1}{2}(4)\Rightarrow 450=V_{C_2H_4}+V_{C_3H_6}+V_{C_4H_8} (5)$
$\Rightarrow (3)-(5)\Rightarrow 560=V_{C_4H_{10}}+V_{C_2H_6}+V_{CH_4} +V_{H_2} (6)$
$\Rightarrow (6)-(5)\Rightarrow V_{C_4H_{10}}=110 lit$
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giải đáp
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Toán Ankan(11).
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$M_Y=2.12=24\Rightarrow M_X=3.24=72\Rightarrow X:C_5H_{12}$
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giải đáp
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Toán Ankan(12).
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$C_\overline{n}H_{2\overline{n}+2}+\frac{3\overline{n}+1}{2}O_2\rightarrow \overline{n}CO_2+(\overline{n}+1)H_2O$
$1 0,5(3\overline{n}+1) $
$n_{O_2}=\frac{54,88}{22,4}=2,45 mol;n_M=\frac{22,2}{14\overline{n}+2}$
$\Rightarrow \frac{1}{0,5(3\overline{n}+1)}=\frac{\frac{22,2}{14\overline{n}+2}}{2,45}\Rightarrow \overline{n}=...................$
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giải đáp
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Toán Ankan(13).
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$C_\overline{n}H_{2\overline{n}+2}+\frac{3\overline{n}+1}{2}O_2\rightarrow \overline{n}CO_2+(\overline{n}+1)H_2O$
$a 0,5a(3\overline{n}+1) \overline{n}a ( \overline{n}+1)a$
$\Rightarrow \begin{cases} \overline{n}a =0,36 \\ 0,5a(3\overline{n}+1)=\frac{63,28.20}{100.22,4} \end{cases}\Rightarrow \begin{cases}a=0,05 \\ \overline{n}=7,2 \end{cases}\Rightarrow m=a(14\overline{n}+2)=5,14 gam$
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