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$1C, 2C , 3B$ 4. $n_{HCl(khí)}$ = $0,5$ mol $m_{dd sau pư}$ = $0,5*36,5 + m = 18,25 + m$ $\sum_{}^{}$ $m_{HCl}$ = $0,5*36,5 + 16*m/100 = 18,25 + 0,16m$ %$HCl$ = $\frac{18,25 + 0,16m}{18,25 + m}$ *$100$% = $21,11$% => $m = 281,75$g 5, $H_2$ + $Cl_2$ => $2HCl$ bd $2 3$ sau $(2-2*0,95 (3 - 2*0,95) (4*0,95)$ => $n_{khí}$ = $2 - 2*0,95 + 3 - 2*0,95) + 4*0,95$ = $5$ lit
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Trả lời 04-03-13 08:00 PM
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