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sửa đổi
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Bài 1
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$d_{Y/H_2}=20\Rightarrow \frac{32n_{O_2}+48n_{O_3}}{n_{O_2}+n_{O_3}}=2.20$$\Rightarrow n_{O_2}=n_{O_3}\Rightarrow 1V_Y\Leftrightarrow 0,5.2+0,5.3=2,5V_Y=V_{[O]}$$CO+[O]\rightarrow CO_2$$H_2+[O]\rightarrow H_2O$$V_X=V_{[O]}=10 lit\Rightarrow V_Y=\frac{10}{2,5}=4 lit$
$d_{Y/H_2}=20\Rightarrow \frac{32n_{O_2}+48n_{O_3}}{n_{O_2}+n_{O_3}}=2.20$$\Rightarrow n_{O_2}=n_{O_3}\Rightarrow 1V_Y\Leftrightarrow V_Y(0,5.2+0,5.3)=2,5V_Y=V_{[O]}$$CO+[O]\rightarrow CO_2$$H_2+[O]\rightarrow H_2O$$V_X=V_{[O]}=10 lit\Rightarrow V_Y=\frac{10}{2,5}=4 lit$
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sửa đổi
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câu khó
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Ta có:$A+H_2O\rightarrow X+Y$$\Rightarrow \begin{cases}A:este \\ X:rượu\\Y:axit \end{cases}$ $*)Y:C_2H_4O_2\Rightarrow Y:CH_3COOH$ $*)X:C_6H_{12}O\Rightarrow X:$ có một liên kết pi (do mạch hở) $X+KMnO_4\rightarrow CH_3-CH_2-CH_2-CH(OH)-CH(OH)-CH_2OH$ $\Rightarrow X:trans-CH_3-CH_2-CH_2-CH=CH-CHOH$ $\Rightarrow A:trans
-CH_3-CH_2-CH_2-CH=CH-CHOOC-CH_3$
Ta có:$A+H_2O\rightarrow X+Y$$\Rightarrow \begin{cases}A:este \\ X:rượu\\Y:axit \end{cases}$ $*)Y:C_2H_4O_2\Rightarrow Y:CH_3COOH$ $*)X:C_6H_{12}O\Rightarrow X:$ có một liên kết pi (do mạch hở) $X+KMnO_4\rightarrow CH_3-CH_2-CH_2-CH(OH)-CH(OH)-CH_2OH$ $\Rightarrow X:trans-CH_3-CH_2-CH_2-CH=CH-CH_2OH$ $\Rightarrow A:trans
-CH_3-CH_2-CH_2-CH=CH-CH_2OOC-CH_3$
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sửa đổi
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BT
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$2Al+Fe_2O_3\rightarrow
2Fe+Al_2O_3 $$2Al+2NaOH+2H_2O\rightarrow 2NaAlO_2+3H_2$ $n_{Al}=\frac{6,48}{27}=0,0888888889 mol;n_{Fe_2O_3}=0,1 mol\Rightarrow H=100\%\Rightarrow Al:$Hết$\Rightarrow n_{H_2}=0,06 mol\Rightarrow n_{Al dư}=0,04 mol\Rightarrow H=\frac{6,48-0,04.27}{6,48}.100=83,33\%$
$2Al+Fe_2O_3\rightarrow
2Fe+Al_2O_3 $$2Al+2NaOH+2H_2O\rightarrow 2NaAlO_2+3H_2$ $n_{Al}=\frac{6,48}{27}=0,24 mol;n_{Fe_2O_3}=0,1 mol\Rightarrow H=100\%\Rightarrow Fe_2O_3:$Hết và $
n_{Al dư}=0,04 $Mặt khác:$ n_{H_2}=0,06 mol\Rightarrow n_{Al dư}=0,04 mol\Rightarrow H=100\%\Rightarrow A:$Là đáp án đúng
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sửa đổi
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BT
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$2Al+Fe_2O_3\rightarrow
2Fe+Al_2O_3 $$2Al+2NaOH+2H_2O\rightarrow 2NaAlO_2+3H_2O$ $n_{Al}=\frac{6,48}{27}=0,0888888889 mol;n_{Fe_2O_3}=0,1 mol\Rightarrow H=100\%\Rightarrow Al:$Hết$\Rightarrow n_{H_2}=0,06 mol\Rightarrow n_{Al dư}=0,04 mol\Rightarrow H=\frac{6,48-0,04.27}{6,48}.100=83,33\%$
$2Al+Fe_2O_3\rightarrow
2Fe+Al_2O_3 $$2Al+2NaOH+2H_2O\rightarrow 2NaAlO_2+3H_2$ $n_{Al}=\frac{6,48}{27}=0,0888888889 mol;n_{Fe_2O_3}=0,1 mol\Rightarrow H=100\%\Rightarrow Al:$Hết$\Rightarrow n_{H_2}=0,06 mol\Rightarrow n_{Al dư}=0,04 mol\Rightarrow H=\frac{6,48-0,04.27}{6,48}.100=83,33\%$
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sửa đổi
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BT
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$2Al+Fe_2O_3\rightarrow
2Fe+Al_2O_3 $$2Al+2NaOH+2H_2O\rightarrow 2NaAlO_2+3H_2O$ $n_{Al}=\frac{6,48}{27}=0,0888888889 mol;n_{Fe_2O_3}=0,1 mol\Rightarrow H=100\%\Rightarrow Al:$Hết$\Rightarrow n_{H_2}=0,06 mol\Rightarrow n_{Al đư}=0,04 mol\Rightarrow H=\frac{6,48-0,04.27}{6,48}.100=83,33\%$
$2Al+Fe_2O_3\rightarrow
2Fe+Al_2O_3 $$2Al+2NaOH+2H_2O\rightarrow 2NaAlO_2+3H_2O$ $n_{Al}=\frac{6,48}{27}=0,0888888889 mol;n_{Fe_2O_3}=0,1 mol\Rightarrow H=100\%\Rightarrow Al:$Hết$\Rightarrow n_{H_2}=0,06 mol\Rightarrow n_{Al dư}=0,04 mol\Rightarrow H=\frac{6,48-0,04.27}{6,48}.100=83,33\%$
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