a) \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\)

            0,3<---0,3<---------0,3<-------0,3

`=>` \(\left\{{}\begin{matrix}m_{Mg}=0,3.24=7,2\left(g\right)\\m_{ZnO}=39,6-7,2=32,4\left(g\right)\end{matrix}\right.\)

b) \(\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{7,2}{39,6}.100\%=18,18\%\\\%m_{ZnO}=100\%-18,18\%=81,82\%\end{matrix}\right.\)

c) \(n_{ZnO}=\dfrac{32,4}{81}=0,4\left(mol\right)\)

PTHH: \(ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\)

              0,4---->0,4

`=>` \(C\%_{H_2SO_4}=\dfrac{\left(0,3+0,4\right).98}{200}.100\%=34,3\%\)

`a)PTHH:`

`2Al + 6HCl -> 2AlCl_3 + 3H_2 \uparrow`

                              `0,1`   `0,15`             `(mol)`

`Cu + HCl -xx->`

`n_[H_2]=[3,36]/[22,4]=0,15(mol)`

`b)m_[AlCl_3]=0,1.133,5=13,35(g)`

\(\begin{array}{l}
a)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
b)\\
{n_{{H_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol\\
{n_{AlC{l_3}}} = 0,15 \times \dfrac{2}{3} = 0,1\,mol\\
{m_{AlC{l_3}}} = 0,1 \times 133,5 = 13,35g
\end{array}\)

Ta có: \(n_{K_2O}=\dfrac{4,7}{94}=0,05\left(mol\right)\)

PTHH: \(K_2O+H_2O\rightarrow2KOH\)

            0,05-------------->0,1

            \(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)

             0,1-------->0,05

`=>` \(C\%_{KOH}=\dfrac{0,1.56}{50}.100\%=11,2\%\)

`=>` \(\left\{{}\begin{matrix}m_{ddH_2SO_4}=\dfrac{0,05.98}{24,5\%}=20\left(g\right)\\V_{ddH_2SO_4}=\dfrac{20}{1,25}=16\left(ml\right)\end{matrix}\right.\)

`200ml=0,2l`

`n_(H_2SO_4` `=0,2.3,25=0,65(mol)`

Gọi `n_{NaOH}=x(mol);n_{BaCl_2}=y(mol)`

PTHH:

`2NaOH+H_2SO_4->Na_2SO_4+2H_2O`

`BaCl_2+H_2SO_4->BaSO_4+2HCl`

Từ pt:

`n_{H_2SO_4}=(1)/(2)n_{NaOH}+n_{BaCl_2}=0,65(mol)`

`=>0,5x+y=0,65`    $(1)$

`n_{Na_2SO_4}=(1)/(2)n_{NaOH}=0,5x(mol)`

`n_{HCl}=2n_{BaCl_2}=2y(mol)`

`=>142.0,5x+36,5.2y=64,7`    $(2)$

$(1)(2)$ `->` $\begin{cases} x=0,5(mol)\\y=0,4(mol) \end{cases}$

`m_(NaOH` `=0,5.40=20(g)`

`m_(BaCl_2` `=0,4.208=83,2(g)`

$n_{Al_2O_3} = \dfrac{10,2}{102} = 0,1(mol)$
$n_{H_2SO_4} = 0,35.1 = 0,35(mol)$
$Al_2O_3 + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2O$

Ta thấy : 

$n_{Al_2O_3} : 1 < n_{H_2SO_4} : 3$ nên $H_2SO_4$ dư

$n_{H_2SO_4\ pư} = 3n_{Al_2O_3} = 0,3(mol) \Rightarrow n_{H_2SO_4\ dư} = 0,35 - 0,3 = 0,05(mol)$

$n_{Al_2(SO_4)_3} = n_{Al_2O_3} = 0,1(mol)$
$C_{M_{H_2SO_4\ dư}} = \dfrac{0,05}{0,35} = 0,14M$
$C_{M_{Al_2(SO_4)_3}} = \dfrac{0,1}{0,35} = 0,28M$

`n_(Al_2O_3` `=(10,2)/(102)=0,1(mol)`

`350ml=0,35l`

`n_(H_2SO_4` `=0,35.1=0,35(mol)`

PTHH: `Al_2O_3+3H_2SO_4->Al_2(SO_4)_3+3H_2O`

Do `n_{Al_2O_3}<(n_{H_2SO_4})/(3)->H_2SO_4` dư

Theo pt:

`n_(Al_2(SO_4)_3` `=n_(Al_2O_3` `=0,1(mol)`

`n_(H_2SO_4(pư)` `=3n_(Al_2O_3` `=0,3(mol)`

`n_(H_2SO_4(dư)` `=0,35-0,3=0,05(mol)`

`C_MAl_2(SO_4)_3` `=(0,1)/(0,35)≈0,29(M)`

`C_MH_2SO_4(dư)` `=(0,05)/(0,35)≈0,14(M)`

Coi $m_{dd\ HCl} = 50(gam)$

$n_{HCl}  = \dfrac{50.14,6\%}{36,5} = 0,2(mol)$
$HCl + NaOH \to NaCl + H_2O$

Theo PTHH : $n_{NaCl} = n_{NaOH} = n_{HCl} = 0,2(mol)$
$m_{dd\ NaOH} = \dfrac{0,2.40}{16\%} = 50(gam)$

$m_{dd\ sau\ pư} = 50 + 50 = 100(gam)$
$C\%_{NaCl} = \dfrac{0,2.58,5}{100}.100\% = 11,7\%$

G/sử `n_{HCl}=1(mol)`

PTHH: `NaOH+HCl->NaCl+H_2O`

Theo pt: `n_(NaOH` `=n_(NaCl` `=n_(HCl` `=1(mol)`

`m_(ddHCl` `=(1.36,5)/(14,6%)=250(g)`

`m_(ddNaOH` `=(1.40)/(16%)=250(g)`

`m_(ddNaCl` `=250+250=500(g)`

`C%_(NaCl` `=(1.58,5)/(500).100%=11,7%`

a)

$Fe + H_2SO_4 \to FeSO_4 + H_2$
$FeO + H_2SO_4 \to FeSO_4 + H_2O$
Theo PTHH : 

$n_{Fe} = n_{H_2}  = \dfrac{2,24}{22,4} = 0,1(mol)$
$m_{Fe} = 0,2.56 = 11,2(gam)$
$m_{FeO} = 18,4 - 11,2 = 7,2(gam)$
b) $n_{FeO} = 0,1(mol)$

Theo PTHH : $n_{H_2SO_4} = n_{Fe} + n_{FeO} = 0,3(mol)$
$m_{dd\ H_2SO_4} = \dfrac{0,3.98}{7\%} = 420(gam)$
c) $m_{dd\ sau\ pư} = 18,4  + 420 - 0,2.2 = 438(gam)$
$n_{FeSO_4} = n_{Fe} + n_{FeO} = 0,3(mol)$
$C\%_{FeSO_4} = \dfrac{0,3.152}{438}.100\% =10,4\%$

$Fe+H_{2}S{O}_4\to FeS{O}_4+H_{2}FeO+H_{2}S{O}_4\to FeS{O}_4+H_{2}O$

nH2=4,48/22,4=0,2(mol)

=>nFe=0,2(mol)=>mFe=0,2.56=11,2(g)

=>mFeO=18,4-11,2=7,2(g)

b)nH2SO4=nH2=0,2(mol)

=>mH2SO4 7%=0,2.98=19,6(g)

=>mH2SO4 =19,6:7%=280(g)

c)mFeSO4=0,2.152=30,4(g)

mdd sau pư=18,4+280-0,2.2=298(g)

=>C%FeSO4=$\frac{30,4}{298}.100\%$=10,2%

a) $n_{K_2O} = \dfrac{9,4}{94} = 0,1(mol)$
$K_2O + H_2O \to 2KOH$
$n_{KOH} = 2n_{K_2O} = 0,2(mol)$
$C_{M_{KOH}} = \dfrac{0,2}{0,2} = 1M$

b)

$2KOH + H_2SO_4 \to K_2SO_4 + 2H_2O$
Ta thấy : 

$n_{KOH} : 2 < n_{H_2SO_4} : 1$ nên $H_2SO_4$ dư

$n_{K_2SO_4} = n_{H_2SO_4\ pư} = \dfrac{1}{2}n_{KOH} = 0,1(mol)$
$n_{H_2SO_4\ dư} = 0,2 - 0,1 = 0,1(mol)$

$m_{dd KOH} = 1,14.200 = 228(gam) ; m_{dd\ H_2SO_4} = 1,12.200 = 224(gam)$
$m_{dd\ sau\ pư} = 228 + 224 = 452(gam)$
$C\%_{K_2SO_4} = \dfrac{0,1.174}{452}.100\% = 3,84\%$

$C\%_{H_2SO_4\ dư} = \dfrac{0,1.98}{452}.100\% = 2,17\%$

a) $n_{K_2SO_4} = 0,2.0,5 = 0,1(mol) ; n_{MgCl_2} = 0,3.0,2 = 0,06(mol)$
$V_{dd\ sau\ khi\ trộn} = 0,2 + 0,3 = 0,5(lít)$
$C_{M_{K_2SO_4}} = \dfrac{0,1}{0,5} = 0,2M$

$C_{M_{MgCl_2}} = \dfrac{0,06}{0,5} = 0,12M$

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