Hóa học

Trung hòa 100ml dung dịch C cần dùng 35ml dung dịch gì thế bn ?

1) \(Al+6HNO_3\rightarrow Al\left(NO_3\right)_3+3NO_2+3H_2O\)

2) \(5Mg+12HNO_3\rightarrow5Mg\left(NO_3\right)_2+N_2+6H_2O\)

3) \(3Cr+27HNO_3\rightarrow8Cr\left(NO_3\right)_3+3NH_3+9H_2O\)

4) \(8Cr+30HNO_3\rightarrow8Cr\left(NO_3\right)_3+3N_2O+15H_2O\)

5) \(10Cr+36HNO_3\rightarrow10Cr\left(NO_3\right)_3+3N_2+18H_2O\)

6) \(4Zn+9HNO_3\rightarrow4Zn\left(NO_3\right)_2+NH_3+3H_2O\)

7) \(4Zn+10HNO_3\rightarrow4Zn\left(NO_3\right)_2+N_2O+5H_2O\)

8) \(8Cu+18HNO_3\rightarrow8Cu\left(NO_3\right)_2+2NH_3+6H_2O\)

9) \(5Cu+12HNO_3\rightarrow5Cu\left(NO_3\right)_2+N_2+6H_2O\)

10) \(3Cu+8HNO_3\rightarrow3Cu\left(NO_3\right)_2+2NO+4H_2O\)

CTHH: `R_2O_3`

Ta có: \(\dfrac{2M_R}{2M_R+16.3}.100\%=70\%\)

=> M_R = 56 (g/mol)

=> R là Fe

- Gọi CTHH của oxit kim loại đó là \(A_2O_3\) (do kim loại đó có hóa trị III).

-Vì kim loại đó chiếm 70% KL oxit.

\(\Rightarrow\)Oxi chiếm 30% KL oxit.

\(m_O=16.3=48\)

\(\Rightarrow m_{A_2O_3}=\dfrac{100.48}{30}=160\)

\(\Rightarrow2M_A+16.3=160\)

\(\Rightarrow M_A=56\).

\(\Rightarrow\)A là Fe (hóa trị III).

a) 

\(Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O\)

\(NaHCO_3+HCl\rightarrow NaCl+CO_2+H_2O\)

b) 

Gọi số mol Na₂CO₃, NaHCO₃ là a, b (mol)

=> 106a + 84b = 19 (1)

Thep PTHH: \(n_{CO_2}=a+b=\dfrac{4,48}{22,4}=0,2\left(mol\right)\) (2)

(1)(2) => a = 0,1 (mol); b = 0,1 (mol)

\(\left\{{}\begin{matrix}m_{Na_2CO_3}=0,1.106=10,6\left(g\right)\\m_{NaHCO_3}=0,1.84=8,4\left(g\right)\end{matrix}\right.\)

c) n_HCl = 2a + b = 0,3 (mol)

=> m_HCl = 0,3.36,5 = 10,95 (g)

=> \(C\%_{dd.HCl}=\dfrac{10,95}{100}.100\%=10,95\%\)

c) n_NaCl = 2a + b = 0,3 (mol)

=> m_NaCl = 0,3.58,5 = 17,55 (g)

m_{dd sau pư} = 100 + 19 - 0,2.44 = 110,2 (g)

=> \(C\%_{NaCl}=\dfrac{17,55}{110,2}.100\%=15,9\%\)

a)

`Na_2CO_3 + 2HCl -> 2NaCl + CO_2 + H_2O`

`NaHCO_3 + HCl -> NaCl + CO_2 + H_2O`

b)

`n_{CO_2} = [4,48]/[22,4] = 0,2 (mol)`

Gọi `n_{Na_2CO_3} = a (mol); n_{NaHCO_3} = b (mol)`

`=> 106a + 84b = 19(1)`

Theo PT: `=> a + b = 0,2(2)`

`(1),(2) => a = b = 0,1`

`m_{Na_2CO_3} = 0,1.106 = 10,6 (g)`

`m_{NaHCO_3} = 0,1.84 = 8,4 (g)`

c)

Theo PT: `n_{HCl} = 2n_{CO_2} = 0,4 (mol)`

`=> C%_{HCl} = [0,4.36,5][100] .100% = 14,6%`

d)

`m_{dd.sau.pư} = 100 + 19 - 0,2.44 = 110,2 (g)`

Theo PT:`n_{NaCl} = 2n_{Na_2CO_3} + n_{NaHCO_3} = 0,3 (mol)`

`=> C%_{NaCl} = [0,3.58,5][110,2] .100% = 15,93%`

1) 

Gọi số mol O₂ sinh ra là a (mol)

=> n_KK = 3a (mol)

\(\left\{{}\begin{matrix}n_{O_2\left(tổng\right)}=a+3a.20\%=1,6a\left(mol\right)\\n_{N_2}=3a.80\%=2,4a\left(mol\right)\end{matrix}\right.\)

\(n_C=\dfrac{0,528}{12}=0,044\left(mol\right)\)

\(m_A=\dfrac{0,894.100}{8,132}=11\left(g\right)\)

TH1: C gồm CO₂, N₂, O₂

Bảo toàn C: \(n_{CO_2}=0,044\left(mol\right)\)

=> \(n_C=\dfrac{0,044.100}{22,92}=\dfrac{110}{573}\left(mol\right)\)

Bảo toàn O: \(n_{O_2\left(C\right)}=1,6a-0,044\left(mol\right)\)

C chứa \(\left\{{}\begin{matrix}CO_2:0,044\left(mol\right)\\O_2:1,6a-0,044\left(mol\right)\\N_2:2,4a\left(mol\right)\end{matrix}\right.\)

=> \(0,044+1,6a-0,044+2,4a=\dfrac{110}{573}\)

=> a = 0,048 (mol)

=> x = \(m_A+m_{O_2}=11+0,048.32=11,536\left(g\right)\)

TH2: C gồm CO₂, CO, N₂

Bảo toàn C: \(n_{CO_2}+n_{CO}=0,044\left(mol\right)\)

Bảo toàn O: \(2.n_{CO_2}+n_{CO}=3,2a\left(mol\right)\)

=> \(n_{CO_2}=3,2a-0,044\left(mol\right)\) 

=> \(n_{hh.C}=\dfrac{\left(3,2a-0,044\right).100}{22,92}=\dfrac{8000}{573}a-\dfrac{110}{573}\left(mol\right)\)

Ta có: \(n_{hh.C}=n_{CO_2}+n_{CO}+n_{N_2}=0,044+2,4a=\dfrac{8000}{573}a-\dfrac{110}{573}\)

=> a = 0,02 (mol)

=> x = 11 + 0,02.32 = 11,64 (g)

2) \(n_{KCl}=\dfrac{0,894}{74,5}=0,012\left(mol\right)\)

PTHH: \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)

               0,008<-------0,012

- Với TH1: \(\%m_{KClO_3}=\dfrac{0,008.122,5}{11,536}.100\%=8,5\%\Rightarrow\%m_{KMnO_4}=100\%-8,5\%=91,5\%\)

- Với TH2:

\(\%m_{KClO_3}=\dfrac{0,008.122,5}{11,64}.100\%=8,42\%\Rightarrow\%m_{KMnO_4}=100\%-8,42\%=91,58\%\)

Gọi \(\left\{{}\begin{matrix}n_X=a\left(mol\right)\\n_Y=2a\left(mol\right)\end{matrix}\right.\)

=> a.M_X +2a.M_Y = 11,7 (1)

\(n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

PTHH: \(X+H_2SO_4\rightarrow XSO_4+H_2\)

            \(2Y+3H_2SO_4\rightarrow Y_2\left(SO_4\right)_3+3H_2\)

Theo PTHH: \(n_{H_2}=n_X+1,5.n_Y=a+3a=0,6\left(mol\right)\)

=> a = 0,15 (mol)

(1) => M_X + 2.M_Y = 78

Chỉ có M_X = 24 (g/mol) ; M_Y = 27 (g/mol)

=> X là Mg, Y là Al

\(m_{NaOH}=0,1.40=4\left(g\right)\)

=> \(C\%=\dfrac{4}{200}.100\%=2\%\)

\(V_{dd}=\dfrac{200}{1,2}=\dfrac{500}{3}\left(ml\right)=\dfrac{1}{6}\left(l\right)\)

=> \(C_M=\dfrac{0,1}{\dfrac{1}{6}}=0,6M\)

\(m_{NaOH}=\dfrac{200.16}{100}=32\left(g\right)\)

=> \(m_{dd.NaOH.25\%}=\dfrac{32.100}{25}=128\left(g\right)\)

=> \(V_{dd.NaOH.25\%}=\dfrac{128}{1,18}=100\left(ml\right)\)

a) Ta có: \(\left\{{}\begin{matrix}\dfrac{p_A}{3}=\dfrac{p_B}{4}=\dfrac{p_C}{5}\\2p_A-p_C=3\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}p_A=9\\p_B=12\\p_C=15\end{matrix}\right.\)

=> A là F, B là Mg, C là P

F chứa 9 hạt proton, 9 hạt electron, 10 hạt nơtron

Mg chứa 12 hạt proton, 12 hạt electron, 12 hạt nơtron

P chứa 15 hạt proton, 15 hạt electron, 16 hạt nơtron

b) 

NTK_A = 9.1 + 10.1 = 19 (đvC)

NTK_B = 12.1 + 12.1 = 24 (đvC)

NTK_C = 15.1 + 16.1 = 31 (đvC)

\(V_{dd.K_3PO_4}=\dfrac{500}{1,25}=400\left(ml\right)=0,4\left(l\right)\\ \Rightarrow n_{K_3PO_4}=0,4.0,5=0,2\left(mol\right)\\ \Rightarrow y=0,2.212=42,4\left(g\right)\)

V_{dd. K3 PO4} = 500/1,25 = 400 ml = 0,4 (l) 

 ==> n_{K3 PO4} = 0,4.0,5 = 0,2 mol

 ==> y = 0,2.212 = 42,4 g