Hoa tan hoan toan 1,04g hon hop Mg,cuo trong 1 luong dd Hcl 7,3% ( vua du cho phan ung) ng ta thu dc 0,224 lit khi (o dieu kien tieu chuan)
a) tinh khoi luong moi chat trong hon hop ban dau?
b) tinh khoi luong d hcl can dung?
c) tinh nong do % cua cac muoi trong dd thu dc sau phan ung?
Rep nhanh e hoc
Đáp án:
\(\begin{array}{l}
a)\\
{m_{Mg}} = 0,24g\\
{m_{CuO}} = 0,8g\\
b)\\
{m_{{\rm{dd}}HCl}} = 20g\\
c)\\
C{\% _{MgC{l_2}}} = 4,51\% \\
C{\% _{CuC{l_2}}} = 6,41\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
CuO + 2HCl \to CuC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{V}{{22,4}} = \dfrac{{0,224}}{{22,4}} = 0,01mol\\
{n_{Mg}} = {n_{{H_2}}} = 0,01mol\\
{m_{Mg}} = n \times M = 0,01 \times 24 = 0,24g\\
{m_{CuO}} = 1,04 – 0,24 = 0,8g\\
b)\\
{n_{CuO}} = \dfrac{m}{M} = \dfrac{{0,8}}{{80}} = 0,01mol\\
{n_{HCl}} = 2{n_M} + 2{n_{CuO}} = 0,04mol\\
{m_{HCl}} = n \times M = 0,04 \times 36,5 = 1,46g\\
{m_{{\rm{dd}}HCl}} = \dfrac{{1,46 \times 100}}{{7,3}} = 20g\\
c)\\
{n_{MgC{l_2}}} = {n_{Mg}} = 0,01mol\\
{m_{MgC{l_2}}} = n \times M = 0,01 \times 95 = 0,95g\\
{n_{CuC{l_2}}} = {n_{CuO}} = 0,01mol\\
{m_{CuC{l_2}}} = n \times M = 0,01 \times 135 = 1,35g\\
{m_{{\rm{dd}}spu}} = 1,04 + 20 – 0,01 \times 2 = 21,02g\\
C{\% _{MgC{l_2}}} = \dfrac{{0,95}}{{21,02}} \times 100\% = 4,51\% \\
C{\% _{CuC{l_2}}} = \dfrac{{1,35}}{{21,02}} \times 100\% = 6,41\%
\end{array}\)