Đáp án:

 a.nZn=6,5/65=0,1 mol

   mHCl=100.14,6/100=14,6 g

   nHCl=14,6/36,5=0,4 mol

           Zn+  2HCl->   ZnCl2+  H2

bài ra: 0,1     0,4

PƯ:      0,1     0,2

Sau PƯ: 0       0,2         0,1        0,1

b. VH2=0,1.22,4=2,24 l

c. mHCl dư=0,2.36,5=7,3 g

   mZnCl2=0,1.136=13,6 g

  mdd sau pư=6,5+100-0,1.2=106,3 g

 C% HCl=7,3.100/106,3=6,87%

C%ZnCl2=13,6.100/106,3=12,79%

Đáp án:

\( {V_{{H_2}}} = 2,24{\text{ lít}}\)

\( C{\% _{ZnC{l_2}}}  = 12,79\% ;C{\% _{HCl}}  = 6,87\% \)

Giải thích các bước giải:

 Phản ứng xảy ra:

\(Zn + 2HCl\xrightarrow{{}}ZnC{l_2} + {H_2}\)

Ta có:

\({n_{Zn}} = \frac{{6,5}}{{65}} = 0,1{\text{ mol}}\)

\({m_{HCl}} = 100.14,6\%  = 14,6{\text{ gam}}\)

\( \to {n_{HCl}} = \frac{{14,6}}{{36,5}} = 0,4{\text{ mol > 2}}{{\text{n}}_{Zn}}\)

Vậy \(HCl\) dư

\( \to {n_{{H_2}}} = {n_{ZnC{l_2}}} = {n_{Zn}} = 0,1{\text{ mol}}\)

\( \to {V_{{H_2}}} = 0,1.22,4 = 2,24{\text{ lít}}\)

\({n_{HCl{\text{ dư}}}} = 0,4 - 0,1.2 = 0,2{\text{ mol}}\)

BTKL:

\({m_{Zn}} + {m_{dd{\text{ HCl}}}} = {m_{dd}} + {m_{{H_2}}}\)

\( \to 6,5 + 100 = {m_{dd}} + 0,1.2 \to {m_{dd}} = 106,3{\text{ gam}}\)

\({m_{ZnC{l_2}}} = 0,1.(65 + 35,5.2) = 13,6{\text{ gam}}\)

\({m_{HCl{\text{ dư}}}} = 0,2.36,5 = 7,3{\text{ gam}}\)

\( \to C{\% _{ZnC{l_2}}} = \frac{{13,6}}{{106,3}} = 12,79\% ;C{\% _{HCl}} = \frac{{7,3}}{{106,3}} = 6,87\% \)