Đáp án:
\(\begin{array}{l}
b)\\
{V_{{H_2}}} = 2,24l\\
c)\\
C{\% _{ZnC{l_2}}} = 6,59\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
b)\\
{n_{Zn}} = \dfrac{{6,5}}{{65}} = 0,1mol\\
{n_{{H_2}}} = {n_{Zn}} = 0,1mol\\
{V_{{H_2}}} = n \times 22,4 = 0,1 \times 22,4 = 2,24l\\
c)\\
{n_{HCl}} = 2{n_{Zn}} = 0,2mol\\
{m_{HCl}} = n \times M = 0,2 \times 36,5 = 7,3g\\
{m_{{\rm{dd}}HCl}} = \dfrac{{7,3 \times 100}}{{3,65}} = 200g\\
{m_{{\rm{dd}}spu}} = 200 + 6,5 - 0,1 \times 2 = 206,3g\\
{n_{ZnC{l_{ 2}}}} = {n_{Zn}} = 0,1mol\\
{m_{ZnC{l_2}}} = n \times M = 0,1 \times 136 = 13,6g\\
C{\% _{ZnC{l_2}}} = \dfrac{{13,6}}{{206,3}} \times 100\% = 6,59\%
\end{array}\)
