$2B+2H_{2}O\rightarrow 2BOH+H_{2}$$0.15$ $0,15$ $0,075$ (mol)
$M_{B}=\frac{5,85}{0,15}=39$ (g/mol)
$\rightarrow B$ là $Kali(K=39)$
$n_{H_{2}}=\frac{1,68}{22,4}=0,075$(mol)
$m_{dd nc}=200.1=200$ (g) (Vì $D_{H_{2}O}=1$ (g/ml))
$m_{dds}=5,85+200-0.075.2=205,7$ (g)
$C$%$=\frac{0,15.56}{205,7}.100=4,08$%
$C_{M}=\frac{0,15}{0,2}=0,75$(M)
Chúc bạn học tốt nhá! ^^