$CO + O_{oxit} \to CO_2$
$CO_2 + Ca(OH)_2 \to CaCO_3 + H_2O$
Ta có : 

$n_{O(oxit)} = n_{CO_2} = n_{CaCO_3} = \dfrac{7}{100} =0,07(mol)$
$\Rightarrow a = m_{Fe} + m_O = 3,92 + 0,07.16 = 5,04(gam)$

\(n_{Fe_2O_3}=\dfrac{3.2}{160}=0.02\left(mol\right)\)

\(n_{HCl}=\dfrac{2.19}{36.5}=0.06\left(mol\right)\)

\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

\(1...........6\)

\(0.02...........0.06\)

Lập tỉ lệ : \(\dfrac{0.02}{1}>\dfrac{0.06}{6}\Rightarrow Fe_2O_3dư\)

\(n_{Fe_2O_3\left(dư\right)}=0.02-\dfrac{0.06}{6}=0.01\left(mol\right)\)

\(m_{Fe_2O_3\left(dư\right)}=0.01\cdot160=1.6\left(g\right)\)

\(m_{FeCl_3}=0.02\cdot162.5=3.25\left(g\right)\)

\(m_{H_2O}=0.03\cdot18=0.54\left(g\right)\)

Đáp án C

2Al + Fe₂O₃ → Al₂O₃ + 2Fe Khi Al dư thu :

Al + NaOH + H₂O →    NaAlO₂ + 1,5 H₂ => n_{Al ban đầu} = 2n_Fe2O3 + 2/3 n _H2 = 0,04 mol => m = 0,04 . 27 = 1,08 g

Đáp án C

Mg+2HCl->MgCl2+H2

0,25--------------------0,25 mol

3H2+Fe2O3-to>2Fe+3H2O

0,25-----1\12-------------1\6 mol

n Mg=6\24=0,25 mol

=>m Fe=1\6 .56=9,33g

=>m Fe2O3=1\12.160=13.33g

\(n_{Mg}=\dfrac{6}{24}=0,25\left(mol\right)\)

PTHH: Mg + 2HCl → MgCl₂ + H₂

Mol:    0,25                               0,25

PTHH: Fe₂O₃ + 3H₂ → 2Fe + 3H₂O

Mol:     1/12        0,25

\(m_{Fe_2O_3}=\dfrac{1}{12}.160=13,33\left(g\right)\)

Đáp án cần chọn là: A

a. PTPƯ : 2Na + 2H2O ---> 2NaOH + H2

+ nH2 = 0,15 mol

+ nNa = 2 nH2 =0,3 mol

=> mNa =n.M= 6,9 g

+ nNaOH = nNa = 0,3 mol

=> mNaOH = n. M= 0,3 . 40= 12 gam

b. Quỳ tím hóa xanh

Mình cần câu c bạn ơi

a, \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

\(FeCl_3+3KOH\rightarrow3KCl+Fe\left(OH\right)_{3\downarrow}\)

\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)

Theo PT: \(n_{HCl}=6n_{Fe_2O_3}=0,6\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,6.36,5=21,9\left(g\right)\)

b, \(n_{Fe\left(OH\right)_3}=n_{FeCl_3}=2n_{Fe_2O_3}=0,2\left(mol\right)\)

\(\Rightarrow m_{Fe\left(OH\right)_3}=0,2.107=21,4\left(g\right)\)

\(n_{KOH}=3n_{FeCl_3}=0,6\left(mol\right)\)

\(\Rightarrow C_{M_{KOH}}=\dfrac{0,6}{0,2}=3\left(M\right)\)