17)

\(FeO\): oxit bazo: sắt (II) oxit

\(HCl\): axit: axit clohidric

\(Al_2O_3\): oxit bazo: nhôm oxit

\(NaOH\): bazo: natri hidroxit

\(N_2O_5\): oxit axit: dinito pentaoxit

\(H_2SO_4\): axit sunfuric

\(CuCl_2\): muối trung tính: đồng (II) clorua

\(NaHCO_3\): muối axit: natri hidrocacbonat

 \(Fe(OH)_3\): bazo: sắt (III) hidroxit

18)

Các phản ứng xảy ra:

\(F{{\text{e}}_3}{O_4} + 4{H_2}\xrightarrow{{{t^o}}}3F{\text{e}} + 4{H_2}O\)

\(2Na + 2{H_2}O\xrightarrow{{}}2NaOH + {H_2}\)

\(2KMn{O_4}\xrightarrow{{{t^o}}}{K_2}Mn{O_4} + Mn{O_2} + {O_2}\)

\(N{a_2}O + {H_2}O\xrightarrow{{}}2NaOH\)

19)

Phản ứng xảy ra:

\(F{\text{e}} + 2HCl\xrightarrow{{}}F{\text{e}}C{l_2} + {H_2}\)

Ta có:

\({m_{HCl}} = 100.3,65\%  = 3,65{\text{ gam}}\)

\( \to {n_{HCl}} = \frac{{3,65}}{{36,5}} = 0,1{\text{ mol}}\)

\( \to {n_{F{\text{e}}}} = {n_{F{\text{e}}C{l_2}}} = {n_{{H_2}}} = \frac{1}{2}{n_{HCl}} = \frac{1}{2}.0,1 = 0,05{\text{ mol}}\)

\( \to m = {m_{Fe}} = 0,05.56 = 2,8{\text{ gam}}\)

\({V_{{H_2}}} = 0,05.22,4 = 1,12{\text{ lít}}\)

\({m_{F{\text{e}}C{l_2}}} = 0,05.(56 + 35,5.2) = 6,35{\text{ gam}}\)

BTKL:

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

\( \to 2,8 + 100 = {m_{dd}} + 0,05.2 \to {m_{dd}} = 102,7{\text{ gam}}\)

\( \to C{\% _{F{\text{e}}C{l_2}}} = \frac{{6,35}}{{102,7}}.100\%  = 6,183\% \)

Đáp án:

 

Giải thích các bước giải:

oxit

FeO sắt 2 oxit

 Al2O3 nhôm oxit

N2O5 nitow oxit

axit 

HCl axit clohidric

H2SO4 ãit sunfuric

bazơ

NaOH natri hidroxit

Fe(OH)3 sắt 3 hidroxit

muối

NaHCO3 natri hidro cacbonat