Đáp án:
a) Natri
b) 1,5M
c) 7,88%
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2R + 2HCl \to 2RCl + {H_2}\\
{n_{{H_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol\\
{n_R} = 0,15 \times 2 = 0,3\,mol\\
{M_R} = \dfrac{{6,9}}{{0,3}} = 23\,g/mol \Rightarrow R:Natri(Na)\\
b)\\
{n_{NaCl}} = {n_{Na}} = 0,3\,mol\\
{C_M}HCl = \dfrac{{0,3}}{{0,2}} = 1,5M\\
c)\\
{m_{{\rm{dd}}HCl}} = 200 \times 1,08 = 216g\\
{m_{{\rm{dd}}spu}} = 6,9 + 216 - 0,15 \times 2 = 222,6g\\
{C_\% }NaCl = \dfrac{{0,3 \times 58,5}}{{222,6}} \times 100\% = 7,88\%
\end{array}\)
