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)$\to x=2,625; y=240g$

$\Rightarrow m_{SO_3}=80x=210g$