cho 5g hon hop CaCO3, CaSO4 vao 20 ml dung dich HCl thu dc 22,4 lit khi. Tinh:
a) Nong do mol dung dich HCl da dung
b) khoi luong moi muoi trong hon hop
c) % ve khoi luong moi muoi trong hon hop
-Mong cac ban tra loi giup minh
cho 5g hon hop CaCO3, CaSO4 vao 20 ml dung dich HCl thu dc 22,4 lit khi. Tinh: a) Nong do mol dung dich HCl da dung b) khoi luong moi muoi trong hon h
By Caroline
Đáp án:
\[CaC{O_3} + 2HCl\xrightarrow{{}}CaC{l_2} + C{O_2} + {H_2}O\]
\[\to {n_{C{O_2}}} = \frac{{0,224}}{{22,4}} = 0,01{\text{ (mol)}}\] \[ \to {{\text{n}}_{HCl}} = 2{n_{C{O_2}}} = 0,01.2 = 0,02{\text{ (mol)}} \] \[\to {{\text{C}}_{M{\text{ HCl}}}} = \frac{{0,02}}{{0,02}} = 1(M)\]
\[{n_{CaC{O_3}}} = {n_{C{O_2}}} = 0,01{\text{( mol)}}\] \[ \to {{\text{m}}_{CaC{O_3}}} = 0,01.100 = 1{\text{ (g)}}\] \[ \to {{\text{m}}_{CaS{O_4}}} = 4{\text{ (g)}}\]
\[\% {m_{CaC{O_3}}} = \frac{1}{5}\times 100\% = 20\% \to \% {m_{CaS{O_4}}} = 100\%-20\%=80\% \]
Đáp án:
a) \({C_{M{\text{ }}HCl}} = 1M\)
b) \({m_{CaC{O_3}}} = 1{\text{ gam; }}{{\text{m}}_{CaS{O_4}}} = 4{\text{ gam}}\)
c) \(\% {m_{CaC{O_3}}} = 20\% ;\% {m_{CaS{O_4}}} = 80\% \)
Giải thích các bước giải:
Đề này thể tích khí sai, 0,224 lít khí thì đúng hơn.
\(CaC{O_3} + 2HCl\xrightarrow{{}}CaC{l_2} + C{O_2} + {H_2}O\)
\(\to {n_{C{O_2}}} = \frac{{0,224}}{{22,4}} = 0,01{\text{ mol}} \to {{\text{n}}_{HCl}} = 2{n_{C{O_2}}} = 0,01.2 = 0,02{\text{ mol}} \to {{\text{C}}_{M{\text{ HCl}}}} = \frac{{0,02}}{{0,02}} = 1M\)
\({n_{CaC{O_3}}} = {n_{C{O_2}}} = 0,01{\text{ mol}} \to {{\text{m}}_{CaC{O_3}}} = 0,01.100 = 1{\text{ gam}} \to {{\text{m}}_{CaS{O_4}}} = 4{\text{ gam}}\)
\(\% {m_{CaC{O_3}}} = \frac{1}{5} = 20\% \to \% {m_{CaS{O_4}}} = 80\% \)