a, \(n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\)

PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)

Theo PT: \(n_{Cu\left(LT\right)}=n_{CuO}=0,2\left(mol\right)\)

\(\Rightarrow m_{Cu\left(LT\right)}=0,2.64=12,8\left(g\right)\)

\(\Rightarrow H=\dfrac{11,25}{12,8}.100\%\approx87,89\%\)

b, \(n_{H_2}=n_{CuO}=0,2\left(mol\right)\)

PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

Theo PT: \(n_{HCl}=2n_{H_2}=0,4\left(mol\right)\)

\(\Rightarrow V_{ddHCl}=\dfrac{0,4}{2}=0,2\left(l\right)=200\left(ml\right)\)

\(\Rightarrow m_{ddHCl}=200.1,2=240\left(g\right)\)

\(n_{MnO_2}=\dfrac{3,48}{87}=0,04(mol)\\ n_{Cl_2}=\dfrac{0,672}{22,4}=0,03(mol)\\ a,MnO_2+4HCl\xrightarrow{t^o}MnCl_2+Cl_2+2H_2O\\ \Rightarrow n_{Cl_2(p/ứ)}=0,04(mol)\\ \Rightarrow V_{Cl_2(p/ứ)}=0,04.22,4=0,896(l)\\ \Rightarrow H\%=\dfrac{0,672}{0,896}.100\%=75\%\)

\(b,n_{Cu}=\dfrac{0,64}{64}=0,01(mol)\\ 2Fe+3Cl_2\xrightarrow{t^o}2FeCl_3(1)\\ Cu+Cl_2\xrightarrow{t^o}CuCl_2(2)\\ \Rightarrow n_{Cl_2(2)}=n_{Cu}=0,01(mol)\\ \Rightarrow n_{Cl_2(1)}=0,03-0,01=0,02(mol)\\ \Rightarrow n_{Fe}=\dfrac{1}{75}(mol) \Rightarrow m_{Fe}=\dfrac{1}{75}.56=0,75(g)\\ \Rightarrow m_{hh}=0,75+0,64=1,39(g)\)

\(c,FeCl_3+3AgNO_3\to Fe(NO_3)_3+3AgCl\downarrow\\ CuCl_2+2AgNO_3\to Cu(NO_3)_2+2AgCl\downarrow\\ \Rightarrow \Sigma n_{AgCl}=3n_{FeCl_3}+2n_{CuCl_2}=0,04+0,01=0,05(mol)\\ \Rightarrow m_{\downarrow}=\Sigma m_{AgCl}=0,05.143,5=7,175(g)\\ d,FeCl_3+3NaOH\to Fe(OH)_3\downarrow+3NaCl\\ CuCl_2+2NaOH\to Cu(OH)_2\downarrow+2NaCl\\ \Rightarrow \Sigma n_{naOH}=3n_{FeCl_3}+2n_{CuCl_2}=0,05(mol)\\ \Rightarrow V_{dd_{NaOH}}=\dfrac{0,05}{2}=0,025(mol)\\ \Rightarrow m_{dd_{NaOH}}=0,025.1,12=0,028(g)\)

a)MnO₂+4HCl→MnCl₂+Cl₂+2H₂O

nMnO₂=3,48/87=0,04mol

n_Cl2=n_MnO2=0,04mol

V_Cl2=0,04×22,4=0,896l

H=0,672/0,896×100%=75%

b)2Fe+3Cl₂→2FeCl₃(1)

Cu+Cl₂→CuCl₂

n_Cu=0,64/64=0,01mol

n_CuCl2=n_Cu=0,01mol

m_CuCl2=0,01×135=1,35g

n_Cl2(1)=0,03−0,01=0,02mol

n_FeCl3=2/3.n_Cl2(1)=1/75mol

m_FeCl3=1/75×162,5=2,17g

m=1,35+2,17=3,52g

F e + 2 H C l → F e C L 2 + H 2

Vì H = 85% nên:

n H 2  = n F e   p ư = 0,01275 mol

V H 2 = 0,01275.22,4= 0,2856 lit

⇒ Chọn C.

\(n_{MnO_2}=\dfrac{17,4}{87}=0,2\left(mol\right)\\ PTHH:MnO_2+4HCl_{đặc,nóng}\rightarrow MnCl_2+Cl_2+2H_2O\\ n_{Cl_2\left(TT\right)}=\dfrac{3,584}{22,4}=0,16\left(mol\right)\\ n_{Cl_2\left(LT\right)}=n_{MnO_2}=0,2\left(mol\right)\\ \Rightarrow H=\dfrac{n_{Cl_2\left(TT\right)}}{n_{Cl_2\left(LT\right)}}.100\%=\dfrac{0,16}{0,2}.100=80\%\)

\(n_{HCl}=\dfrac{150.36,5\%}{36,5}=1,5\left(mol\right)\)

PTHH: MnO₂ + 4HCl --> MnCl₂ + Cl₂ + 2H₂O

                           1,5------------->0,375

=> V_Cl2 = 0,375.22,4 = 8,4 (l)

PT: \(2KMnO_4+16HCl\rightarrow2MnCl_2+2KCl+5Cl_2+8H_2O\)

Ta có: \(n_{KMnO_4}=\dfrac{31,6}{158}=0,2\left(mol\right)\)

\(m_{HCl}=\dfrac{60.36,5}{100}=21,9\left(g\right)\Rightarrow n_{HCl}=\dfrac{21,9}{36,5}=0,6\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,2}{2}>\dfrac{0,6}{16}\), ta được KMnO₄ dư.

Theo PT: \(n_{Cl_2\left(LT\right)}=\dfrac{5}{16}n_{HCl}=0,1875\left(mol\right)\)

Mà: H% = 80%

\(\Rightarrow n_{Cl_2\left(TT\right)}=0,1875.80\%=0,15\left(mol\right)\)

\(\Rightarrow V_{Cl_2}=0,15.22,4=3,36\left(l\right)\)

Bạn tham khảo nhé!

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

Cứ 1 mol CaCO3------->1 mol CO2

                    100g------->22,4 l

                     50g--------->11,2L

mà H%=85%------>VCO2=11,2.85/100=9,52(l)

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

\(K_2CO_3+2HCl\rightarrow2KCl+CO_2+H_2O\)

Ta có: 106n_Na2CO3 + 138n_K2CO3 = 43,5 (1)

Theo PT: \(n_{CO_2}=n_{Na_2CO_3}+n_{K_2CO_3}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Na_2CO_3}=0,15\left(mol\right)\\n_{K_2CO_3}=0,2\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Na_2CO_3}=\dfrac{0,15.106}{43,5}.100\%\approx36,55\%\\\%m_{K_2CO_3}\approx63,45\%\end{matrix}\right.\)

b, \(n_{HCl\left(pư\right)}=2n_{CO_2}=0,7\left(mol\right)\)

⇒ n_{HCl (dư)} = 0,7.15% = 0,105 (mol)

\(\Rightarrow V_{ddHCl}=\dfrac{0,7+0,105}{2}=0,4025\left(M\right)\)

c, \(\left\{{}\begin{matrix}n_{NaCl}=2n_{Na_2CO_3}=0,3\left(mol\right)\\n_{KCl}=2n_{K_2CO_3}=0,4\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}C_{M_{NaCl}}=\dfrac{0,3}{0,4025}\approx0,745\left(M\right)\\C_{M_{KCl}}=\dfrac{0,4}{0,4025}\approx0,994\left(M\right)\end{matrix}\right.\)

MnO\(_2\)+4HCl\(\rightarrow\)MnCl\(_2\)+Cl\(_2\)+2H\(_2O\)

0,45 0,45 (mol)

n\(_{MnO_2}\)=\(\dfrac{39,15}{87}\)=0,45(mol)

2Fe + 3Cl\(_2\)\(\rightarrow\)2FeCl\(_3\)

0,3 0,45 0,3 (mol)

m\(_{FeCl_3}\)=0,3.162,5=48,75(g)

vì hiệu suất phản ứng là 86% nên:

m\(_{FeCl_3}\)=\(\dfrac{86.48,75}{100}\)=41,925(g)

2/

Mg+Cl\(_2\)\(\rightarrow\)MnCl\(_2\)

0,6 0,6

n\(_{Mg}\)=\(\dfrac{14,4}{24}\)=0,6(mol)

2\(KMnO_4+16HCl\rightarrow2MnCl_2+2KCl+5Cl_2\uparrow+8H_2O\)

0,24 0,6

vì hiệu suất phản ứng bằng 80%,nên để điều chế 0,6 mol Cl\(_2\)thì cần số mol \(KMnO_4\) là:

n\(_{KMnO_4}\)=\(\dfrac{0,24.100}{80}\)=0,3(mol)

m\(KMnO_4\)=0,3.158=47,4(g)

3/

2

0,4 1 (mol)

=\(\dfrac{63,2}{158}\)=0,4(mol)

V\(_{Cl_2}\)=1.22,4=22,4(l)

vì Hiệu suất phản ứng là 75% nên:

V\(_{Cl_2}\)=\(\dfrac{75.22,4}{100}\)=16,8(l)