a)2NaOH+H2SO4→Na2SO4+2H2O(1)

Cu(NO3)2+2NaOH→Cu(OH)2+2NaNO3(2)

Cu(OH)2→CuO+H2O(3)

nCuO=\(\dfrac{1,6}{80}\)=0,02mol

mddNaOH=31,25×1,12=35g

nNaOH=35×16%40=0,14mol

nNaOH(2)=0,02×2=0,04mol

⇒nNaOH(1)=0,14−0,04=0,1mol

nH2SO4=0,12=0,05mol

CM(H2SO4)=\(\dfrac{0,05}{0,05}\)=1M

CM(Cu(NO3)2)=\(\dfrac{0,02}{0,05}\)=0,4M

b)nCu=\(\dfrac{2,4}{64}\)=0,0375mol

nH+=2nH2SO4=0,1mol

nNO3−=2nCu(NO3)2=0,04mol

Cu+4H++NO3−→Cu2++NO+2H2O

\(\dfrac{0,04}{1}\)>\(\dfrac{0,03751}{1}\)>\(\dfrac{0,1}{4}\)⇒ Tính theo ion H+nNO=0,14=0,025mol

⇒VNO=0,025×22,4=0,56l

cho cac axit :HCLO,HNO3,H2S,H2SO3,HNO2,HCLO4,HMno4.so axit manh la

\(3NaOH+FeCl_3\rightarrow Fe\left(OH\right)_3+3NaCl\)

\(n_{NaCl}=n_{NaOH}=0,2.3=0,6\left(mol\right)\)

=> \(C_{M\left(NaCl\right)}=\dfrac{0,6}{0,2}=3M\)

\(n_{Fe\left(ỌH\right)_3}=\dfrac{1}{3}n_{NaOH}=0,2\left(mol\right)\)

\(2Fe\left(OH\right)_3-^{t^o}\rightarrow Fe_2O_3+3H_2O\)

Ta có \(n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe\left(OH\right)_3}=0,1\left(mol\right)\)

=> m _Fe2O3 = 0,1 . 160=16(g)

FeCl2+ 2NaOH --> Fe(OH)2 + 2NaCl (1)

4Fe(OH)2 +O2 --t^o-> 2Fe2O3 + 4H2O (2)

nFeCl2=0,2(mol)

nNaOH=0,5(mol)

Lập tỉ lệ :

\(\dfrac{0,2}{1}< \dfrac{0,5}{2}\)

=> FeCl2 hết ,NaOH dư

Theo (1,2) : nFe2O3=1/2nFeCl2=0,1(mol)

=> x=16(g)

b) V_NaOH=\(\dfrac{m}{D}=\dfrac{200}{1,12}\approx178,6\left(ml\right)\)\(\approx\)0,1786(l)

Theo (1) : nNaOH(PƯ)=2nFeCl2=0,4(mol)

=>nNaOH dư=0,1(mol)

nNaCl=2nFeCl2=0,4(mol)

=> CM dd NaCl\(\approx\)2,24(M)

C_{M dd NaOH dư}\(\approx\)0,6(M)

a)

$Fe +H_2SO_4 \to FeSO_4 + H_2$

$FeSO_4 + 2KOH \to Fe(OH)_2 + K_2SO_4$
$4Fe(OH)_2 + O_2 \xrightarrow{t^o} 2Fe_2O_3 + 4H_2O$

$n_{Fe_2O_3} = \dfrac{20}{160} = 0,125(mol)$

Theo PTHH : $n_{Fe} = 2n_{Fe_2O_3} = 0,25(mol)$
$m_{Fe} = 0,25.56 = 14(gam)$

b)

$n_{H_2} = n_{Fe} = 0,25(mol)$
$V_{H_2} = 0,25.22,4 = 5,6(lít)$
c)

$n_{H_2SO_4} = n_{Fe} = 0,25(mol)$
$V_{dd\ H_2SO_4} = \dfrac{0,25}{1} = 0,25(lít) = 250(ml)$

\(PTHH:Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ FeSO_4+2KOH\rightarrow Fe\left(OH\right)_2+K_2SO_4\\4 Fe\left(OH\right)_2+O_2\underrightarrow{t^o}2Fe_2O_3+4H_2O\)

\(a.n_{Fe_2O_3}=\dfrac{20}{160}=0,125\left(mol\right)\\ n_{H_2}=n_{H_2SO_4}=n_{Fe}=n_{FeSO_4}=n_{Fe\left(OH\right)_2}=\dfrac{4}{2}.0,125=0,25\left(mol\right)\\ m_{Fe}=0,25.56=14\left(g\right)\\ b.V_{H_2\left(đktc\right)}=0,25.22,4=5,6\left(l\right)\\ c.V_{ddH_2SO_4}=\dfrac{0,25}{1}=0,25\left(l\right)=250\left(ml\right)\)

Đề không đề cập nung trong điều kiện nào nên mình coi như nung trong không khí nhé.

PT: \(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)

\(MgSO_4+2NaOH\rightarrow Mg\left(OH\right)_2+Na_2SO_4\)

\(FeSO_4+2NaOH\rightarrow Fe\left(OH\right)_2+Na_2SO_4\)

\(Cu\left(OH\right)_2\underrightarrow{t^o}CuO+H_2O\)

\(Mg\left(OH\right)_2\underrightarrow{t^o}MgO+H_2O\)

\(4Fe\left(OH\right)_2+O_2\underrightarrow{t^o}2Fe_2O_3+4H_2O\)

Giả sử dd chứa a (l)

Ta có: n_CuSO4 = 0,2a (mol), n_MgSO4 = 0,1a (mol), n_FeSO4 = 0,2a (mol)

Theo PT: \(\left\{{}\begin{matrix}n_{CuO}=n_{Cu\left(OH\right)_2}=n_{CuSO_4}=0,2a\left(mol\right)\\n_{MgO}=n_{Mg\left(OH\right)_2}=n_{MgSO_4}=0,1a\left(mol\right)\\n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe\left(OH\right)_2}=\dfrac{1}{2}n_{FeSO_4}=0,1a\left(mol\right)\end{matrix}\right.\)

⇒ 0,2a.80 + 0,1a.40 + 0,1a.160 = 18

⇒ a = 0,5 (l)

⇒ V = 500 (ml)

Bạn ơi , câu 1 sao dd NaCl lại tác dụng với NaCl v ?