5.
$m_{H_2SO_4}=100.10\%+50.40\%=30g$
$\Rightarrow C\%_{H_2SO_4}=\dfrac{30.100}{100+50}=20\%$
6.
$n_{K_2O}=\dfrac{188}{94}=2 mol$
$K_2O+H_2O\to 2KOH$
$\Rightarrow n_{KOH}=2.2=4 mol$
$m_{KOH}=4.56=224g$
$\Rightarrow m_{dd}=224:5,6\%=4000g$
$\Rightarrow m_{H_2O}=4000-188=3812g$
7.
Gọi x(g) là khối lượng dd 7,93%.
$\Rightarrow m_{KOH(dd)}=0,0793x(g)$
$n_{K_2O}=\dfrac{47}{94}=0,5 mol$
$\Rightarrow n_{KOH}=0,5.2=1 mol$
$\Sigma m_{KOH}=0,0793x+56(g)$
$m_{dd}=x+47(g)$
$\Rightarrow 0,0793x+56=0,21(x+47)$
$\Leftrightarrow x=352,95(g)$
8.
Gọi x là mol $SO_3$, y là $m_{dd}$ 49%.
$\Rightarrow 80x+y=450$ (1)
$SO_3+H_2O\to H_2SO_4$
$\Rightarrow \Sigma m_{H_2SO_4}=0,49y+98x$
$\Rightarrow 98x+0,49y=450.83,3\%$ (2)
(1)(2)$\to x=2,625; y=240g$
$\Rightarrow m_{SO_3}=80x=210g$
