Đáp án:
\({V_{{O_2}}} = 42{\text{ lít}}\)
\({V_{kk}} = 210{\text{ lít}}\)
\({m_{C{O_2}}} = 66{\text{ gam}}\)
\({m_{{H_2}O}} = 13,5{\text{ gam}}\)
\({m_{CaC{O_3}}} = 150{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(2{C_2}{H_2} + 5{O_2}\xrightarrow{{{t^o}}}4C{O_2} + 2{H_2}O\)
Ta có:
\({V_{{O_2}}} = \frac{5}{2}{V_{{C_2}{H_2}}} = \frac{5}{2}.16,8 = 42{\text{ lít}}\)
\( \to {V_{kk}} = 5{V_{{O_2}}} = 42.5 = 210{\text{ lít}}\)
Ta có:
\({n_{{C_2}{H_2}}} = \frac{{16,8}}{{22,4}} = 0,75{\text{ mol}}\)
\( \to {n_{C{O_2}}} = 2{n_{{C_2}{H_2}}} = 1,5{\text{ mol;}}{{\text{n}}_{{H_2}O}} = {n_{{C_2}{H_2}}} = 0,75{\text{ mol}}\)
\( \to {m_{C{O_2}}} = 1,5.44 = 66{\text{ gam}}\)
\({m_{{H_2}O}} = 0,75.18 = 13,5{\text{ gam}}\)
\(Ca{(OH)_2} + C{O_2}\xrightarrow{{}}CaC{O_3} + {H_2}O\)
\({n_{CaC{O_3}}} = {n_{C{O_2}}} = 1,5{\text{ mol}}\)
\( \to {m_{CaC{O_3}}} = 1,5.100 = 150{\text{ gam}}\)