Đáp án:

b) 3,08l

c) 20,075 g

Giải thích các bước giải:

a)2Cu+O2→2CuO

2Fe+O2→2FeO

4Fe+3O2→2Fe2O3

3Fe+2O2→Fe3O4

CuO+2HCl→CuCl2+H2O

FeO+2HCl→FeCl2+H2O

Fe2O3+6HCl→2FeCl3+3H2

Fe3O4+8HCl→2FeCl3+FeCl2+4H2O

b)mO2=mhhX−mhh=19,2−14,8=4,4g

nO2=4,432=0,1375mol

VO2=0,1375×22,4=3,08l

c)nO=0,1375×2=0,275mol

nHCl=2nO=0,55mol

mHCl=0,55×36,5=20,075g

Đáp án:

\({V_{{O_2}}} = 3,36{\text{ lít}}\)

\( {m_{HCl}} = 21,9{\text{ gam}}\)

Giải thích các bước giải:

 Các phản ứng xảy ra:

\(2Cu + {O_2}\xrightarrow{{{t^o}}}2CuO\)

\(3Fe + 2{O_2}\xrightarrow{{{t^o}}}F{e_3}{O_4}\)

\(2Fe + {O_2}\xrightarrow{{{t^o}}}2FeO\)

\(4Fe + 3{O_2}\xrightarrow{{{t^o}}}2F{e_2}{O_3}\)

\(CuO + 2HCl\xrightarrow{{}}CuC{l_2} + {H_2}O\)

\(FeO + 2HCl\xrightarrow{{}}FeC{l_2} + {H_2}O\)

\(F{e_2}{O_3} + 6HCl\xrightarrow{{}}2FeC{l_3} + 3{H_2}O\)

\(F{e_3}{O_4} + 8HCl\xrightarrow{{}}FeC{l_2} + 2FeC{l_3} + 4{H_2}O\)

BTKL:

\({m_{kl}} + {m_{{O_2}}} = {m_X}\)

\( \to 14,8 + {m_{{O_2}}} = 19,2 \to {m_{{O_2}}} = 4,8{\text{ gam}}\)

\( \to {n_{{O_2}}} = \frac{{4,8}}{{16.2}} = 0,15{\text{ mol}}\)

\( \to {V_{{O_2}}} = 0,15.22,4 = 3,36{\text{ lít}}\)

Bảo toàn \(O\):

\({n_{{H_2}O}} = 2{n_{{O_2}}} = 0,15.2 = 0,3{\text{ mol}}\)

Bảo toàn \(H\)

\({n_{HCl}} = 2{n_{{H_2}O}} = 0,3.2 = 0,6{\text{ mol}}\)

\( \to {m_{HCl}} = 0,6.36,5 = 21,9{\text{ gam}}\)