Đáp án:
\({m_{CaC{O_3}}} = 48{\text{ gam}}\)
\( V= 16,128{\text{ lít}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(Ca{(OH)_2} + S{O_2}\xrightarrow{{}}CaS{O_3} + {H_2}O\)
\(Ca{(OH)_2} + 2S{O_2}\xrightarrow{{}}Ca{(HS{O_3})_2}\)
\(Ca{(HS{O_3})_2} + 2NaOH\xrightarrow{{}}CaS{O_3} + N{a_2}S{O_3} + 3{H_2}O\)
Ta có:
\({n_{CaC{O_3}{\text{ mới}}}} = \frac{{12}}{{100}} = 0,12{\text{ mol = }}{{\text{n}}_{Ca{{(HC{O_3})}_2}}}\)
\({n_{Ca{{(OH)}_2}}} = 0,6.1 = 0,6{\text{ mol}} \to {{\text{n}}_{CaC{O_3}}} = 0,6 - 0,12 = 0,48{\text{ mol}}\)
\( \to {m_{CaC{O_3}}} = 0,48.100 = 48{\text{ gam}}\)
\({n_{C{O_2}}} = {n_{CaC{O_3}}} + 2{n_{Ca{{(HC{O_3})}_2}}} = 0,48 + 0,12.2 = 0,72{\text{ mol}}\)
\( \to {V_{C{O_2}}} = 0,72.22,4 = 16,128{\text{ lít}}\)
