Đáp án:

$ \dfrac{1600}{37} gam$

Giải thích các bước giải:

Gọi $n_{SO_3} = x (mol)$

$\rightarrow m_{SO_3} = 80x (g)$

PTHH: $SO_3 + H_2O \rightarrow H_2SO_4$

Theo PT $n_{H_2SO_4} = n_{SO_3} = x (mol)$

$\rightarrow m_{H_2SO_4} = \dfrac{200.10}{100} + 98x = 20 + 98x (g)$

$m_{dd} = 200 + 80x (g)$

$\rightarrow \dfrac{20 + 98x}{200+80x} .100\% = 30\%$

$\Leftrightarrow x = \dfrac{20}{37} (mol)$

$\rightarrow m_{SO_3} = \dfrac{20}{37} . 80 = \dfrac{1600}{37} (g)$

$\text{#Kudo/Shinichi}$

\(n_{SO_3^{}}^{} = x\;{\text{mol}}\)

\({\text{m}}_{SO_3^{}}^{} = 80x(g)\)

\(m_{H_2^{}SO_4^{}}^{} = 200.10\%  = 20g\)

\(SO_3^{} + H_2^{}O \to H_2^{}SO_4^{}\)

\(x\,{\text{           x           x}}\)

\(m_{dd_{sauPU}^{}}^{} = 80x + 200\)

\(C\% _{H_2^{}SO_4^{}}^{} = 30\%  = \frac{{20 + 98x}}{{80x + 200}}.100\)

\( \to x = \frac{{20}}{{37}} \to m_{SO_3^{}}^{} = \frac{{20}}{{37}}.80 = \frac{{1600}}{{37}}\)