Đáp án:
\(\begin{array}{l}
a)\\
\% Zn = 61,61\% \\
\% ZnO = 38,39\% \\
b)\\
C{\% _{ZnC{l_2}}} = 3,04\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
ZnO + 2HCl \to ZnC{l_2} + {H_2}O\\
{n_{{H_2}}} = \dfrac{V}{{22,4}} = \dfrac{{2,24}}{{22,4}} = 0,1mol\\
{n_{Zn}} = {n_{{H_2}}} = 0,1mol\\
\Rightarrow {m_{Zn}} = n \times M = 0,1 \times 65 = 6,5g\\
{m_{ZnO}} = 10,55 - 6,5 = 4,05g\\
\% Zn = \dfrac{{{m_{Zn}}}}{m} \times 100\% = \dfrac{{6,5}}{{10,55}} \times 100\% = 61,61\% \\
\% ZnO = 100 - 61,61 = 38,39\% \\
b)\\
{n_{ZnO}} = \dfrac{m}{M} = \dfrac{{4,05}}{{81}} = 0,05mol\\
{n_{HCl}} = 2{n_{Zn}} + 2{n_{ZnO}} = 0,3mol\\
{V_{HCl}} = \dfrac{n}{{{C_M}}} = \dfrac{{0,3}}{{0,5}} = 0,6l = 600ml\\
{m_{{\rm{dd}}HCl}} = d \times V = 1,1 \times 600 = 660g\\
{m_{{\rm{dd}}spu}} = m + {m_{{\rm{dd}}HCl}} - {m_{{H_2}}} = 10,55 + 660 - 0,1 \times 2 = 670,35g\\
{n_{ZnC{l_2}}} = {n_{Zn}} + {n_{ZnO}} = 0,15mol\\
{m_{ZnC{l_2}}} = n \times M = 0,15 \times 136 = 20,4g\\
C{\% _{ZnC{l_2}}} = \dfrac{{20,4}}{{670,35}} \times 100\% = 3,04\%
\end{array}\)
