nHCl=0,7.1=0,7(mol)

nCO2=4,48/22,4=0,2(mol)

PTHH: MgCO3 +2 HCl -> MgCl2 + CO2 + H2O

0,2_________0,4_____0,2______0,2(mol)

nMgCO3=nCO2=0,2(mol) => mMgCO3=0,2. 84= 16,8(g)

=> mFeO= mX - mMgCO3= 24 - 16,8= 7,2(g)

=> %mMgCO3= (16,8/24).100=70%

=>%mFeO=100% - 70%= 30%

b)  nFeO= 7,2/72=0,1(mol)

FeO +2 HCl -> FeCl2 + H2O

0,1___0,2______0,1(mol)

Vì 0,4+0,2=0,6 => HCl có dư => nHCl(dư)= 0,7 - 0,6=0,1(mol)

Vddsau= VddHCl=0,7(l)

CMddHCl(dư)= 0,1/0,7= 1/7 (M)

CMddFeCl2= 0,1/0,7=1/7(M)

CMddMgCl2= 0,2/0,7=2/7(M)

a. \(n_{CO_2}=0,2\left(mol\right)\)

\(MgCO_3+2HCl\rightarrow MgCl_2+H_2O+CO_2\)

\(\Rightarrow n_{MgCl_2}=n_{MgCO_3}=n_{CO_2}=0.2\left(mol\right)\)

\(\Rightarrow m_{MgCO_3}=0,2.84=16,8\left(g\right)\)

\(\%m_{MgCO_3}=\dfrac{16,8.100\%}{24}=70\%\\ \%m_{FeO}=100\%-70\%=30\%\)

b. \(FeO+2HCl\rightarrow FeCl_2+H_2O\)

\(n_{FeO}=\dfrac{24-16,8}{72}=0,1\left(mol\right)=n_{FeCl_2}\)

\(C_{M_{FeCl_2}}=\dfrac{0,1}{0,7}=\dfrac{1}{7}\left(M\right)\)

\(C_{M_{MgCO_3}}=\dfrac{0,2}{0,7}=\dfrac{2}{7}\left(M\right)\)

a) \(FeO+2HCl\rightarrow FeCl_2+H_2\)

\(MgCO_3+2HCl\rightarrow MgCl_2+H_2O+CO_2\)

\(n_{MgCO_3}=n_{CO_2}=0,2\left(mol\right)\)

=> \(\%m_{MgCO_3}=\dfrac{0,2.84}{24}.100=70\%\)

=> %FeO = 30%

b) \(n_{FeO}=\dfrac{24-16,8}{72}=0,1\left(mol\right)\)

\(n_{HCl\left(bđ\right)}=0,7\left(mol\right)\)

\(n_{HCl\left(pứ\right)}=0,1.2+0,2.2=0,6\left(mol\right)\)

=> \(n_{HCl\left(dư\right)}=0,7-0,6=0,1\left(mol\right)\)

Dung dịch Y gồm HCl dư (0,1) , FeCl2 (0,1) , MgCl2 (0,2)

=> \(CM_{HCl}=\dfrac{0,1}{0,7}=0,14M\)

\(CM_{FeCl_2}=\dfrac{0,1}{0,7}=0,14M\)

\(CM_{MgCl_2}=\dfrac{0,2}{0,7}=0,29M\)

a) \(2Fe\left(OH\right)_3-^{t^o}\rightarrow Fe_2O_3+3H_2O\)

\(Cu\left(OH\right)_2-^{t^o}\rightarrow CuO+H_2O\)

Gọi x,y lần lượt là số mol Fe(OH)3 và Cu(OH)2 

=> \(\left\{{}\begin{matrix}107x+98y=20,5\\160.\dfrac{x}{2}+80y=16\end{matrix}\right.\)

=> x= 0,1 ; y=0,1

=> \(\%m_{Fe\left(OH\right)_3}=\dfrac{0,1.107}{20,5}.100=52,2\%\)

\(\%m_{Cu\left(OH\right)_2}=47,8\%\)

b) \(2Fe\left(OH\right)_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+6H_2O\)

\(Cu\left(OH\right)_2+H_2SO_4\rightarrow CuSO_4+2H_2O\)

\(n_{H_2SO_4}=0,1.\dfrac{3}{2}+0,1=0,25\left(mol\right)\)

\(m_{ddH_2SO_4}=\dfrac{0,25.98}{20\%}=122,5\left(g\right)\)

\(m_{ddsaupu}=20,5+122,5=143\left(g\right)\)

\(C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{0,05.400}{143}.100=13,97\%\)

\(C\%_{CuSO_4}=\dfrac{0,1.160}{143}.100=11,19\%\)

c) \(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)

\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

\(n_{Fe_2O_3}=0,05\left(mol\right);n_{CuO}=0,1\left(mol\right)\)

=> \(n_{H_2SO_4}=0,05.3+0,1=0,25\left(mol\right)\)

\(m_{ddH_2SO_4\left(pứ\right)}=\dfrac{0,25.98}{20\%}=122,5\left(g\right)\)

=> \(m_{ddH_2SO_4\left(bđ\right)}=122,5.110\%=134,75\left(g\right)\)

a) \(n_{CO_2}=\dfrac{0,4958}{24,79}=0,02\left(mol\right);n_{HCl}=0,6.1=0,6\left(mol\right)\)

PTHH: \(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)

            0,02<------0,04<----0,02<-----0,02

\(\Rightarrow n_{HCl\left(p\text{ư}\right)}< n_{HCl\left(b\text{đ}\right)}\left(0,04< 0,6\right)\Rightarrow HCl\) dư, \(CaCO_3\) tan hết

\(\Rightarrow\left\{{}\begin{matrix}m_{CaCO_3}=0,02.100=2\left(g\right)\\m_{CaSO_4}=5-2=3\left(g\right)\end{matrix}\right.\)

b) dd sau phản ứng có: \(\left\{{}\begin{matrix}n_{HCl\left(d\text{ư}\right)}=0,6-0,04=0,56\left(mol\right)\\n_{CaCl_2}=0,02\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}C_{M\left(HCl\left(d\text{ư}\right)\right)}=\dfrac{0,56}{0,6}=\dfrac{14}{15}M\\C_{M\left(CaCl_2\right)}=\dfrac{0,02}{0,6}=\dfrac{1}{30}M\end{matrix}\right.\)

nHCl=0,6 mol

FeO+2HCl-->FeCl2+ H2O

x mol               x mol

Fe2O3+6HCl-->2FeCl3+3H2O

x mol                   2x mol

72x+160x=11,6         =>x=0,05 mol

A/ C_FeCl2=0,05/0,3=1/6 M

CFeCl3=0,1/0,3=1/3 M

CHCl du=(0,6-0,4)/0,3=2/3 M

B/ 

NaOH+ HCl-->NaCl+H2O

0,2          0,2

2NaOH+FeCl2-->2NaCl+Fe(OH)2

0,1           0,05

3NaOH+FeCl3-->3NaCl+Fe(OH)3

0,3            0,1

nNaOH=0,6

CNaOH=0,6/1,5=0,4M

Câu 1 bạn biết làm k chỉ mình với

Gọi CT của oxit KL là M2Om 
=> %M = 2M/(2M + 16m) = 85.22% 
<=> M = 46.13m --> ko có KLoại quen thuộc (chỉ có m=2, M = 92.26 ~ Nb = 92.9) 
Tuy nhiên, ta ko cần tìm M mà vẫn tính dc (nhưng bạn vẫn nên xem lại đề nhé)
M2Om + mH2SO4 ---> M2(SO4)m + mH2O 
n(M2Om) = 10/(2M + 16m) = 10/(2*46.13m + 16m) = 10/108.26m 
--> nH2SO4 = m*10/108.26m = 10/108.26 ~ 0.0924 mol 
=> mddH2SO4 = 0.0924*98/0.1 = 90.55g

a)Gọi x,y lần lượt là số mol của Al, Fe trong hỗn hợp ban đầu (x,y>0)

Sau phản ứng hỗn hợp muối khan gồm: \(\left\{{}\begin{matrix}AlCl_3:x\left(mol\right)\\FeCl_2:y\left(mol\right)\end{matrix}\right.\)

Ta có hệ phương trình: \(\left\{{}\begin{matrix}27x+56y=13,9\\133,5x+127y=38\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\approx0,0896\\y\approx0,205\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,0896\cdot27\cdot100\%}{13,9}\approx17,4\%\\\%m_{Fe}=\dfrac{0,205\cdot56\cdot100\%}{13,9}\approx82,6\%\end{matrix}\right.\)

Theo Bảo toàn nguyên tố Cl, H ta có:\(n_{H_2}=\dfrac{n_{HCl}}{2}=\dfrac{3n_{AlCl_3}+2n_{FeCl_2}}{2}\\ =\dfrac{3\cdot0,0896+2\cdot0,205}{2}=0,3394mol\\ \Rightarrow V_{H_2}=0,3394\cdot22,4\approx7,6l\)

 \(n_{CO2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

PTHH : \(CaO+2HCl\rightarrow CaCl_2+H_2O\) (1)

\(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\) (2)

a) Theo Pt : \(n_{CO2}=n_{CaCO3}=0,1\left(mol\right)\)

\(m_{CaCO3}=0,1100=10\left(g\right)\)

\(m_{CaO}=12,8-10=2,8\left(g\right)\)

b) Chắc tính V của dd HCl đã dùng 

(1) \(n_{CaO}=\dfrac{2,8}{56}=0,05\left(mol\right)\) , \(n_{HCl}=2n_{CaO}=0,1\left(mol\right)\)

(2) \(n_{HCl}=2n_{CaCO3}=0,2\left(mol\right)\)

\(\Rightarrow V_{ddHCl}=\dfrac{0,1+0,2}{1}=0,3\left(l\right)=300\left(ml\right)\)