a)
Gọi \({n_{S{O_3}}} = x{\text{ mol}}\)
Ta có: \({m_{{H_2}S{O_4}}} = 500.22,5\% = 112,5{\text{ gam}}\)
\(S{O_3} + {H_2}O\xrightarrow{{}}{H_2}S{O_4}\)
Ta có:
\({m_{{H_2}S{O_4}trong{\text{ }}dung{\text{ }}dịch{\text{mới}}}} = 112,5 + 98x;{m_{dd{\text{ mới}}}} = 500 + 80x \to \frac{{112,5 + 98x}}{{500 + 80x}} = 42,5\% \to x = 1,5625{\text{ gam}} \to {{\text{m}}_{S{O_3}}} = 1,5625.80 = 125{\text{ gam}}\)
b)
BTKL: \({V_1}.1,05 + {V_2}.1,12 = 2000.1,1 = 2200{\text{ gam}}\)
\({m_{NaOH}} = 2200.8\% = {V_1}.1,05.3\% + {V_2}.1,12.10\% \)
Giải được:
\({V_1} = 598,64ml;{V_2} = 1403ml\)