Giải thích các bước giải:

1,

\(\begin{array}{l}
4P + 5{O_2} \to 2{P_2}{O_5}\\
{n_P} = 0,1mol\\
 \to {n_{{O_2}}} = \dfrac{5}{4}{n_P} = 0,125mol\\
 \to {V_{{O_2}}} = 2,8l\\
 \to {V_{KK}} = 5{V_{{O_2}}} = 14l\\
 \to {n_{{P_2}{O_5}}} = \dfrac{1}{2}{n_P} = 0,05mol \to {m_{{P_2}{O_5}}} = 7,1g
\end{array}\)

2,

\(\begin{array}{l}
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
{n_{Zn}} = 0,2mol\\
 \to {n_{{H_2}}} = {n_{Zn}} = 0,2mol \to {V_{{H_2}}} = 4,48l\\
 \to {n_{HCl}} = 2{n_{Zn}} = 0,4mol \to {m_{HCl}} = 14,6g\\
 \to C\% HCl = \dfrac{{14,6}}{{200}} \times 100\%  = 7,3\% \\
 \to {n_{ZnC{l_2}}} = {n_{Zn}} = 0,2mol \to {m_{ZnC{l_2}}} = 27,2g\\
 \to {m_{{\rm{dd}}}} = {m_{Zn}} + {m_{HCl{\rm{dd}}}} - {m_{{H_2}}} = 13 + 200 - 0,2 \times 2 = 212,6g\\
 \to C\% ZnC{l_2} = \dfrac{{27,2}}{{212,6}} \times 100\%  = 12,79\% \\
F{e_2}{O_3} + 3{H_2} \to 2Fe + 3{H_2}O\\
{n_{F{e_2}{O_3}}} = 0,05mol\\
\dfrac{{{n_{{H_2}}}}}{3} > {n_{F{e_2}{O_3}}}
\end{array}\)

\({H_2}\) dư

\( \to {n_{Fe}} = 2{n_{F{e_2}{O_3}}} = 0,1mol \to {m_{Fe}} = 5,6g\)

3,

\(\begin{array}{l}
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
{n_{{H_2}}} = 0,2mol\\
 \to {n_{{H_2}S{O_4}}} = {n_{{H_2}}} = 0,2mol\\
 \to C{M_{{H_2}S{O_4}}} = \dfrac{n}{V} = 1M\\
{H_2} + \dfrac{1}{2}{O_2} \to {H_2}O\\
{n_{{O_2}}} = 0,15mol\\
{n_{{H_2}}} < \dfrac{{{n_{{O_2}}}}}{{0,5}}
\end{array}\)

\({O_2}\) dư

\( \to {n_{{H_2}O}} = {n_{{H_2}}} = 0,2mol \to {m_{{H_2}O}} = 3,6g\)

bài 1:

ta có phương trình:4P+5O2=>2P2O5

a,ta có mP=3.1(g)=>nP=$\frac{3.1}{31}$=0.1(mol) 

=>nO2=$\frac{0.1*5}{4}$=0.125(mol) 

=>vO2=0.125*22.4=2.8(lít)

b,vKK=2.8*5=14 lít

c,nP2O5=$\frac{0.1*2}{4}$=0.05(mol) 

=>mP2O5=0.05*(31*2+16*5)=7.1(g)

bài 2:

ta có phương trình:Zn+2HCl=>ZnCl2+H2

a,ta có mZn=13(g)=>nZn=$\frac{13}{65}$=0.2(mol) 

=>nH2=0.2(mol)=>vH2=0.2*22.4=4.48(lít)

b,ta có nHCl=0.2*2=0.4(mol)

=>mHCl=0.4*36.5=14.6(g)

=>C%=$\frac{14.6}{200}$*100=7.3% 

c,m dd ZnCl2=13+14.6=27.6(g)

nZnCl2=0.2(mol)=>mZnCl2=0.2*(65+35.5*2)=27.2(g)

=>C%=$\frac{27.2}{27.6}$*100=98.5% 

d,Fe2O3+3H2=>2Fe+3H2O

mFe2O3=8(g)=>nFe2O3=$\frac{8}{56*2+16*3}$=0.05(mol) 

ta có tỉ lệ:nH2:nFe2O3=$\frac{0.2}{3}$>$\frac{0.05}{1}$(h2 dư,Fe2O3 hết) 

=>nfe=0.05*2=0.1(mol)=>mFe=0.1*56=5.6(g)

bài 3:

ta có phương trình:Fe+H2SO4=>FeSO4+H2

a,ta có vH2=4.48(lít)=>nH2=$\frac{4.48}{22.4}$=0.2(mol) 

=>$C_{M}$= $\frac{0.2}{0.2}$=1M 

b,2H2+o2=>2H2O

vO2=3.36(lít)=>nO2=$\frac{3.36}{22.4}$=0.15(mol) 

ta có tỉ lệ:nH2:nO2=$\frac{0.2}{2}$<$\frac{0.15}{1}$(H2 hết,O2 dư) 

=>nH2O=nH2=0.2(mol)

=>mH2O=0.2*18=3.6(g)