\(\left\{{}\begin{matrix}FeS_2:x\left(mol\right)\\Cu_2S:y\left(mol\right)\end{matrix}\right.\)

4FeS₂ + 11O₂ \(\xrightarrow{t^o}\) 2Fe₂O₃ + 8SO₂

x..............................0,5x.........2x...................(mol)

Cu₂S + 2O₂ \(\xrightarrow{t^o}\) 2CuO + SO₂

y..........................2y...........y.............................(mol)

Fe₂O₃ + 3H₂SO₄ → Fe₂(SO₄)₃ + 3H₂O

0,5x...........1,5x..................................................(mol)

CuO + H₂SO₄ → CuSO₄ + H₂O

2y............2y.................................................(mol)

Suy ra: 1,5x + 2y = 0,2.2,5(1)

SO₂ + 2NaOH → Na₂SO₃ + H₂O

(2x+y).....(4x+2y).........................................(mol)

Suy ra : 4x + 2y = 0,25.4(2)

Từ (1)(2) suy ra x = 0,2 ; y = 0,1

Vậy :

\(\%m_{FeS_2} = \dfrac{0,2.120}{0,2.120+0,1.160}.100\% = 60\%\\ \%m_{Cu_2S} = 100\% -60\% = 40\%\)

a) 2Al + 6HCl -> 2AlCl3 + 3H2

Al2O3 + 6HCl -> 2AlCl3 + 3H2O

nH2 = 0,15mol => nAl=0,1mol => mAl=2,7g; mAl2O3 = 10,2g => nAl2O3 = 0,1mol

=>%mAl=20,93% =>%mAl2O3 = 79,07%

b) nHCl = 0,1.3+0,1.6=0,9 mol=>mHCl(dd)=100g

mddY=12,9+100-0,15.2=112,6g

mAlCl3=22,5g=>C%=19,98%

a) n_NaOH = 0,6.1 = 0,6 (mol)

PTHH: NaOH + CH₃COOH --> CH₃COONa + H₂O

               0,6----->0,6

=> m_CH3COOH = 0,6.60 = 36 (g)

=> m_C2H5OH = 45,2 - 36 = 9,2 (g)

b) \(n_{C_2H_5OH}=\dfrac{9,2}{46}=0,2\left(mol\right)\)

PTHH: 2CH₃COOH + 2Na --> 2CH₃COONa + H₂

                   0,6---------------------------->0,3

           2C₂H₅OH + 2Na --> 2C₂H₅ONa + H₂

                  0,2--------------------------->0,1

=> V = (0,3 + 0,1).22,4 = 8,96 (l)

C32: 

a, \(n_C=\dfrac{2,4}{12}=0,2\left(mol\right)\)

PT: \(C+O_2\underrightarrow{t^o}CO_2\)

Theo PT: \(n_{CO_2}=n_C=0,2\left(mol\right)\Rightarrow V_{CO_2}=0,2.22,4=4,48\left(l\right)\)

b, \(n_{NaOH}=0,3.1=0,3\left(mol\right)\)

\(\Rightarrow\dfrac{n_{NaOH}}{n_{CO_2}}=1,5\) → Pư tạo NaHCO₃ và Na₂CO₃

PT: \(CO_2+NaOH\rightarrow NaHCO_3\)

\(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)

Ta có: \(\left\{{}\begin{matrix}n_{CO_2}=n_{NaHCO_3}+n_{Na_2CO_3}=0,2\\n_{NaOH}=n_{NaHCO_3}+2n_{Na_2CO_3}=0,3\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}n_{NaHCO_3}=0,1\left(mol\right)\\n_{Na_2CO_3}=0,1\left(mol\right)\end{matrix}\right.\)

⇒ m_NaHCO3 = 0,1.84 = 8,4 (g)

m_Na2CO3 = 0,1.106 = 10,6 (g)

c, \(C_{M_{NaHCO_3}}=C_{M_{Na_2CO_3}}=\dfrac{0,1}{0,3}=\dfrac{1}{3}\left(M\right)\)

Lần sau bạn đăng tách câu hỏi ra nhé.

C31:

a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(ZnO+2HCl\rightarrow ZnCl_2+H_2O\)

b, \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

Theo PT: \(n_{Zn}=n_{H_2}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{0,1.65}{14,6}.100\%\approx44,52\%\\\%m_{ZnO}\approx55,48\%\end{matrix}\right.\)

c, \(n_{ZnO}=\dfrac{14,6-0,1.65}{81}=0,1\left(mol\right)\)

Theo PT: \(n_{HCl}=2n_{Zn}+2n_{ZnO}=0,4\left(mol\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{0,4.36,5}{10\%}=146\left(g\right)\)

Câu 1:

a) CaCO₃ + 2HCl --> CaCl₂ + CO₂ + H₂O

CaO + 2HCl --> CaCl₂ + H₂O

b) 

\(n_{CO_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

PTHH: CaCO₃ + 2HCl --> CaCl₂ + CO₂ + H₂O

_____0,4<---------0,8<-----------------0,4

=> m_CaCO3 = 0,4.100 = 40(g)

=> m_CaO = 62,4 - 40 = 22,4 (g)

c) \(n_{CaO}=\dfrac{22,4}{56}=0,4\left(mol\right)\)

CaO + 2HCl --> CaCl₂ + H₂O

_0,4-->0,8

=> n_HCl = 0,8 + 0,8 = 1,6(mol)

=> \(C_{M\left(HCl\right)}=\dfrac{1,6}{0,25}=6,4M\)

Câu 1:

\(a,CaO+2HCl\to CaCl_2+H_2O\\ CaCO_3+2HCl\to CaCl_2+H_2O+CO_2\uparrow\\ b,n_{CO_2}=\dfrac{8,96}{22,4}=0,4(mol)\\ \Rightarrow n_{CaCO_3}=0,4(mol)\\ \Rightarrow m_{CaCO_3}=0,4.100=40(g)\\ \Rightarrow m_{CaO}=62,4-40=22,4(g)\\ c,n_{CaO}=\dfrac{22,4}{56}=0,4(mol)\\ \Rightarrow \Sigma n_{HCl}=0,4.2+0,4.2=1,6(mol)\\ \Rightarrow C_{M_{HCl}}=\dfrac{1,6}{0,25}=6,4M\)

Câu 2: Đề thiếu