Đáp án
\(H = 83,33\% \)
\(\% {m_{CaO}} = 73,68\% ;\% {m_{CaC{O_3}{\text{ dư}}}} = 26,32\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(CaC{O_3}\xrightarrow{{{t^o}}}CaO + C{O_2}\)
BTKL: \({m_{CaC{O_3}}} = {m_A} + {m_{C{O_2}}}\)
\( \to {m_{C{O_2}}} = 12 - 7,6 = 4,4{\text{ gam}}\)
\( \to {n_{C{O_2}}} = \frac{{4,4}}{{44}} = 0,1{\text{ mol = }}{{\text{n}}_{CaC{O_3}{\text{ phản ứng}}}}\)
\({n_{CaC{O_3}{\text{ ban đầu}}}} = \frac{{12}}{{100}} = 0,12{\text{ mol}}\)
Hiệu suất: \(H = \frac{{0,1}}{{0,12}} = 83,33\% \)
Trong A chứa \(CaO; CaCO_3 dư\)
\({n_{CaO}} = {n_{C{O_2}}} = 0,1 \to {m_{CaO}} = 0,1.56 = 5,6{\text{ gam}}\)
\( \to \% {m_{CaO}} = \frac{{5,6}}{{7,6}} = 73,68\% \to \% {m_{CaC{O_3}{\text{ dư}}}} = 26,32\% \)
Hòa tan hoàn toàn A trong HCl dư.
\(CaO + 2HCl\xrightarrow{{}}CaC{l_2} + {H_2}O\)
\(CaC{O_3} + 2HCl\xrightarrow{{}}CaC{l_2} + C{O_2} + {H_2}O\)
Ta có: \({n_{C{O_2}}} = {n_{CaC{O_3}{\text{ dư}}}} = 0,12 - 0,1 = 0,02{\text{ mol}}\)
\({n_{NaOH}} = 0,125.0,2 = 0,025{\text{ mol}}\)
\( \to 1 < \frac{{{n_{NaOH}}}}{{{n_{C{O_2}}}}} = \frac{{0,025}}{{0,02}} = 1,25 < 2\)
Do vậy tạo hỗn hợp 2 muối.
\(2NaOH + C{O_2}\xrightarrow{{}}N{a_2}C{O_3} + {H_2}O\)
\(NaOH + C{O_2}\xrightarrow{{}}NaHC{O_3}\)
Ta có:
\({n_{N{a_2}C{O_3}}} = {n_{NaOH}} - {n_{C{O_2}}} = 0,005{\text{ mol}}\)
\({n_{NaHC{O_3}}} = 0,02 - 0,005 = 0,015{\text{ mol}}\)
\( \to {C_{M{\text{ N}}{{\text{a}}_2}C{O_3}}} = \frac{{0,005}}{{0,125}} = 0,04M\)
\({C_{M{\text{ NaHC}}{{\text{O}}_3}}} = \frac{{0,015}}{{0,125}} = 0,12M\)
Cho B vào \(Ca(OH)_2\) dư
\(N{a_2}C{O_3} + Ca{(OH)_2}\xrightarrow{{}}CaC{O_3} + 2NaOH\)
\(Ca{(OH)_2} + NaHC{O_3}\xrightarrow{{}}CaC{O_3} + NaOH + {H_2}O\)
\( \to {n_{CaC{O_3}}} = {n_{N{a_2}C{O_3}}} + {n_{NaHC{O_3}}}\)
\( = 0,005 + 0,015 = 0,02{\text{ mol}}\)
\( \to {m_{CaC{O_3}}} = 0,02.100 = 2{\text{ gam}}\)
