Câu 1:

PTHH:

$S{O_3} + {H_2}O \to {H_2}S{O_4}$

${m_{dd{H_2}S{O_4}}} = 500.1,2 = 600g \Rightarrow {m_{{H_2}S{O_4}(bd)}} = \dfrac{{600.24,5}}{{100}} = 147g$

Đặt ${n_{S{O_3}}} = x$

$ \Rightarrow $Sau khi trộn ta có: ${m_{{H_2}S{O_4}(sau)}} = 98x + 147;{m_{ddsau}} = 80x + 600$

$ \Rightarrow C{\% _{{H_2}S{O_4}}} = \dfrac{{98x + 147}}{{80x + 600}}.100\%  = 49\%  \Rightarrow x = 2,5$

$ \Rightarrow {m_{S{O_3}}} = 2,5.80 = 200g$

Câu 2:

PTHH:

$S{O_3} + {H_2}O \to {H_2}S{O_4}$

${n_{S{O_3}}} = 2,5mol \Rightarrow {n_{{H_2}S{O_4}}} = 2,5mol \Rightarrow {m_{{H_2}S{O_4}}} = 245g$

${m_{{H_2}S{O_4}(bd)}} = 0,49m$

$ \Rightarrow $Sau khi trộn ta có: ${m_{{H_2}S{O_4}(sau)}} = 245 + 0,49m;{m_{ddsau}} = 200 + m$

$ \Rightarrow C{\% _{{H_2}S{O_4}}} = \dfrac{{245 + 0,49m}}{{200 + m}}.100\%  = 78,4\%  \Rightarrow m = 300$