(Câu 11, 12 em xem sgk nha)
Câu 9:
$n_{Ba}=\frac{13,7}{137}=0,1 mol$
$n_{H_2SO_4}=\frac{75.4,9\%}{98}= 0,0375 mol$
$Ba+H_2SO_4\to BaSO_4+H_2$
$\Rightarrow n_{H_2SO_4 pứ}= n_{BaSO_4}= n_{H_2}= 0,0375 mol$
$n_{Ba dư}=0,1-0,0375= 0,0625 mol$
$Ba+2H_2O\to Ba(OH)_2+H_2$
$\Rightarrow n_{Ba(OH)_2}= n_{H_2}= 0,0625 mol$
$m_{dd spứ}= m_{Ba}+ 49-m_{BaSO_4}-m_{H_2}$
$= 13,7+49-0,0375.233-2.(0,0375+0,0625)= 53,7625g$
$\Rightarrow C\%_{Ba(OH)_2}=\frac{0,0625.171.100}{53,7625}=19,88\%$
Câu 10:
a,
$n_{H_2}=\frac{11,2}{22,4}=0,5 mol$
Gọi a, b là mol Mg, Al.
$\Rightarrow 24a+27b=10,2$ (1)
$Mg+2HCl\to MgCl_2+H_2$
$2Al+6HCl\to 2AlCl_3+3H_2$
$\Rightarrow a+1,5b=0,5$ (2)
(1)(2) $\Rightarrow a=b=0,2$
$\%m_{Mg}=\frac{0,2.24.100}{10,2}= 47\%$
$\%m_{Al}=53\%$
b,
$n_{MgCl_2}= n_{AlCl_3}=0,2 mol$
$m_{dd spứ}= 10,2+200-0,5.2=209,2g$
$C\%_{MgCl_2}=\frac{0,2.95.100}{209,2}=9\%$
$C\%_{AlCl_3}=\frac{133,5.0,2.100}{209,2}= 12,8\%$
