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$