Đáp án C

Trung hòa phần 1 vừa đủ bởi NaOH:

nAl(OH)3= 15.6/78=0.2 mol

PI : nH2SO4= 0.2 mol

2Al(OH)3 + 3H2SO4 --> Al2(SO4)3 + 6H2O

Bđ: 0.2_________0.2

Pư : 1/15________0.2_________1/15

Kt: 2/15_________0___________1/15

mAl2(SO4)3= 1/15*342= 22.8g

PII:

nNaOH = 0.05 mol

NaOH + Al(OH)3 --> NaAlO2 + 2H2O

Bđ: 0.05_____0.2

Pư: 0.05_____0.05______0.05

Kt: 0________0.15______0.05

mNaAlO2= 0.05*82=4.1g

a) \(n_{H_2SO_4}=0,2\times1=0,2\left(mol\right)\)

\(m_{Al\left(OH\right)_3}=\frac{1}{2}\times15,6=7,8\left(g\right)\)

\(\Rightarrow n_{Al\left(OH\right)_3}=\frac{7,8}{78}=0,1\left(mol\right)\)

PTHH: 2Al(OH)₃ + 3H₂SO₄ → Al₂(SO₄)₃ + 6H₂O

Theo PT: \(n_{Al\left(OH\right)_3}=\frac{2}{3}n_{H_2SO_4}\)

Theo bài: \(n_{Al\left(OH\right)_3}=\frac{1}{2}n_{H_2SO_4}\)

Vì \(\frac{1}{2}< \frac{2}{3}\) ⇒ H₂SO₄ dư

Theo PT: \(n_{Al_2\left(SO_4\right)_3}=\frac{1}{2}n_{Al}=\frac{1}{2}\times0,1=0,05\left(mol\right)\)

\(\Rightarrow m_{Al_2\left(SO_4\right)_3}=0,05\times342=17,1\left(g\right)\)

b) \(n_{NaOH}=0,05\times1=0,05\left(mol\right)\)

\(m_{Al\left(OH\right)_3}=\frac{1}{2}\times15,6=7,8\left(g\right)\)

\(\Rightarrow n_{Al\left(OH\right)_3}=\frac{7,8}{78}=0,1\left(mol\right)\)

PTHH: Al(OH)₃ + NaOH → NaAlO₂ + 2H₂O

Theo PT: \(n_{Al\left(OH\right)_3}=n_{NaOH}\)

Theo bài : \(n_{Al\left(OH\right)_3}=2n_{NaOH}\)

Vì \(2>1\) → Al(OH)₃ dư

Theo Pt: \(n_{NaAlO_2}=n_{NaOH}=0,05\left(mol\right)\)

\(\Rightarrow m_{NaAlO_2}=0,05\times82=4,1\left(g\right)\)

\(PTHH:Na_2CO_3+H_2SO_4\rightarrow Na_2SO_4+H_2O+CO_2\\ CO_2+KOH\rightarrow KHCO_3\\ \Rightarrow n_{H_2SO_4}=n_{CO_2}=1\cdot0,2=0,2\left(mol\right)\\ m_{KOH}=\dfrac{98\cdot40\%}{100\%}=39,2\left(g\right)\\ \Rightarrow n_{KOH}=\dfrac{39,2}{56}=0,7\left(mol\right)\)

Do đó sau phản ứng, KOH dư

\(\Rightarrow n_{KHCO_3}=n_{CO_2}=0,2\left(mol\right)\\ \Rightarrow m_{KHCO_3}=0,2\cdot100=20\left(g\right)\)

a, \(Na_2SO_3+H_2SO_4\rightarrow Na_2SO_4+SO_2+H_2O\)

\(K_2SO_3+H_2SO_4\rightarrow K_2SO_4+SO_2+H_2O\)

Ta có: \(n_{SO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

Theo PT: \(n_{H_2SO_4}=n_{SO_2}=0,3\left(mol\right)\)

\(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,3.98}{20\%}=147\left(g\right)\)

b, Ta có: 126n_Na2SO3 + 158n_K2SO3 = 44,2 (1)

Theo PT: \(n_{SO_2}=n_{Na_2SO_3}+n_{K_2SO_3}=0,3\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Na_2SO_3}=0,1\left(mol\right)\\n_{K_2SO_3}=0,2\left(mol\right)\end{matrix}\right.\)

Có: m _{dd sau pư} = 44,2 + 147 - 0,3.64 = 172 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{Na_2SO_3}=\dfrac{0,1.126}{172}.100\%\approx7,33\%\\C\%_{K_2SO_3}=\dfrac{0,2.158}{172}.100\%\approx18,37\%\end{matrix}\right.\)

c, \(n_{Ba\left(OH\right)_2}=0,5.1=0,5\left(mol\right)\)

\(\Rightarrow\dfrac{n_{SO_2}}{n_{Ba\left(OH\right)_2}}=0,6< 1\) → Pư tạo BaSO₃.

PT: \(SO_2+Ba\left(OH\right)_2\rightarrow BaSO_3+H_2O\)

\(n_{BaSO_3}=n_{SO_2}=0,3\left(mol\right)\Rightarrow m_{BaSO_3}=0,3.217=65,1\left(g\right)\)

a) n Fe = a(mol) ; n Cu = b(mol)

=> 56a + 64b = 4,8(1)

n SO2 = 2,24/22,4 = 0,1(mol)

Bảo toàn e :

3n Fe + 2n Cu = 2n SO2

<=> 3a + 2b = 0,2(2)

Từ (1)(2) suy ra a = b = 0,04

%m Fe = 0,04.56/4,8  .100% = 46,67%

%m Cu = 100% -46,67% = 53,33%

b)

n KOH = 0,15

Ta có :

1 < n KOH / n SO2 = 0,15/0,1 = 1,5 < 2 nên muối sinh ra là Na2SO3(x mol) và NaHSO3(y mol)

2NaOH + SO2 $\to$ Na2SO3 + H2O

NaOH + SO2 $\to$ NaHSO3

Theo PTHH :

n SO2 = a + b = 0,1

n NaOH = 2a + b = 0,15

Suy ra a= 0,05 ; b = 0,05

m muối = 0,05.126 + 0,05.104 = 11,5 gam

Bài 2 :

- Gọi số mol Fe và Cu lần lượt là a, b mol

Ta có : mhh = mFe + mCu = 56a + 64b = 4,8

Bte : 3a + 2b = 0,2 

=> a = b =0,04mol

a, Ta có : mFe =n.M = 2,24g ( 46,7% )

=> %Cu = 53,3%

b, Ta thấy sau phản ứng thu được K+, SO3-2 xmol, HSO3- y mol

BtS : x + y = 0,1 

BTĐT : 2x + y = 0,15 

=> x =y = 0,05 mol

=> mM = mK2SO3 + mKHSO3 = 13,9g

Chọn đáp án C

Đáp án A

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{ZnCl_2}=n_{Zn}=n_{H_2}=0,15\left(mol\right);n_{HCl}=2.0,15=0,3\left(mol\right)\\ a,m=m_{Zn}=0,15.65=9,75\left(g\right)\\ b,C_{MddHCl}=\dfrac{0,3}{0,15}=0,2\left(l\right)\\ c,m_{ZnCl_2}=0,15.136=20,4\left(g\right)\)