Pt tác dụng H₂SO₄ loãng

CuO + H₂SO₄ \(\rightarrow\) CuSO₄ + H₂O (1)
Cu không tác dụng. 
Cu + 2H₂SO₄đặc,n \(\rightarrow\) CuSO₄ + SO₂ + 2H₂O (2)
n_SO2= \(\frac{1,12}{22,4}\)  = 0,05 mol

\(\rightarrow\) n_Cu= n_SO2= 0,05 mol 

% Cu = \(\frac{0,05x64}{10}.100\%\)= 32%

\(\rightarrow\) % CuO = 68%.

Giải tiếp:

\(n_{KMnO_4\left(1\right)}=0,2;n_{KClO_3}=0,25\left(mol\right)\)

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

0,25____________0,375

\(2KMnO_4\overset{t^o}{\rightarrow}K_2MnO_4+MnO_2+O_2\)

0,15____________________________0,075

\(\Rightarrow n_{KMnO_4\left(\text{dư}\right)}=0,05\left(mol\right)\Rightarrow\%KMnO_4\left(Y\right)=16,52\%\)

Đặt \(n_{KMnO_4}=x;n_{KClO_3}=y\left(mol\right)\)

\(\Rightarrow158x+122,5y=62,225\left(1\right)\)

BTKL \(\Rightarrow m_{O_2}=14,4\left(g\right)\Rightarrow n_{O_2}=0,45\left(mol\right)\)

BT e \(\Rightarrow2n_{Cl_2}+4n_{O_2}=5n_{KMnO_4}+6n_{KClO_2}=5x+6y\left(mol\right)\)

\(\Rightarrow2n_{Cl_2}=5x+6y-1,8\left(mol\right)\)

\(BTNT.Mn\Rightarrow n_{MnCl_2}=n_{KMnO_4}=x\left(mol\right)\\ BTNT.K\Rightarrow n_{KCl}=n_{KMnO_4}+n_{KClO_3}=x+y\left(mol\right)\\ BTNT.Cl\Rightarrow n_{KClO_3}+n_{HCl}=n_{KCl}+2n_{MnCl_2}+2n_{Cl_2}\\ \Rightarrow y+1,3=x+y+2x+5x+6y-1,8\left(2\right)\)

\(\left(1\right);\left(2\right)\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,25\end{matrix}\right.\)

\(\Rightarrow m_{KMnO_4}=31,6\left(g\right)\Rightarrow\%KMnO_4=50,78\%\)

giả sử có 100g KMnO4, KClO3

⇒m chất rắn =75g

2KMnO4\(\underrightarrow{^{to}}\)K2MnO4+MnO2+O2

x___________x/2_____ x /2__ x/2 (mol)

2KClO3→2KCl+3O2

y _________y___3y/2

\(\left\{{}\begin{matrix}\text{158x+122.5y=100}\\\text{142x+74.5y=75 }\end{matrix}\right.\rightarrow\left\{{}\begin{matrix}\text{x=0.3}\\\text{y=0.4 }\end{matrix}\right.\)

⇒mKMnO4⇒%

⇒mKClO3⇒%

Zn + 2HCl => ZnCl2 + H2

Na2CO3 + 2HCl=> 2NaCl + H2O + CO2

M_Y = 0,5875.32 = 18,8

áp dụng sơ đồ đường chéo ta đc nH2 : nCO2 = 3:2

mà nH2 = nZn ; nCO2 = nNa2CO3

=> nZn = 3/2 nCO2

ta có \(65.\frac{3}{2}x+106x=4,07\left(g\right)\) => x= 0,02 mol => nZn =0,03

a. => % na2CO3 = \(\frac{0,02.106}{4,07}.100\%=52,088\%\)

=> % Zn = 47,912%

b. nHCl pư = 2 .nZn + 2. nNa2CO3 = 2.0,03+ 2.0,02 = 0,1

=> mHCl pư = 0,1.36,5 = 3,65 (g)

=> m HCl dùng = 3,65.120% = 4,38 (g)

=> mdd HCl = \(\frac{4,38.100}{25}=17,52\)

=> mdd = 4,07 + 17,52 - 0,03.2-0,02.44 = 20,65(g)

mHCl dư = 4,38 - 3,65 = 0,73(g)

C% HCl dư = \(\frac{0,73}{20,65}.100\%\) = 3,535%