Pthh: 

Mg+ H²SO⁴(loãng)-> MgSO⁴ + H²

Fe + H²SO⁴(loãng)-> FeSO⁴ +H²

Mọi người giúp mình vs

a) nco2=v/22.4=0.1 mol

500ml=0.5l

=> nNaoh=Cm.v=0.2 . 0.5=0.1 mol

lập tỉ lệ:

nNaoh/nCo2=0.1/0.1=1

=> sảy ra phương trình

Naoh + co2 ->nahco3

mCo2=n.M=4.4 (g)

mNaoh=n.M=4 (g)

adđlbtkl ta có

mCo2 + mNaoh = mNahco3

=>mNahco3=8.4 (g)

Cau 1 : 

\(n_{H2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

Pt : \(Fe+2HCl\rightarrow FeCl_2+H_2|\)

        1          2             1         1 

       0,1                                 0,1

a) Chat trong dung dich A thu duoc la : sat (II) clorua

    Chat ran B la : dong

    Chat khi C la : khi hidro

b) \(n_{Fe}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)

\(m_{Fe}=0,1.56=5,6\left(g\right)\)

\(m_{Cu}=10-5,6=4,4\left(g\right)\)

⁰/₀Fe = \(\dfrac{5,6.100}{10}=56\)⁰/₀

⁰/₀Cu = \(\dfrac{4,4.100}{10}=44\)⁰/₀

c) Co : \(m_{Cu}=4,4\left(g\right)\)

\(n_{Cu}=\dfrac{4,4}{64}=0,06875\left(mol\right)\)

Pt : \(Cu+2H_2SO_{4dac}\underrightarrow{t^o}CuSO_4+SO_2+2H_2O|\)

        1           2                  1              1          2

     0,06875                                    0,06875

\(n_{SO2}=\dfrac{0,06875.1}{1}=0,06875\left(mol\right)\)

\(V_{SO2\left(dktc\right)}=1,54\left(l\right)\)

 Chuc ban hoc tot

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)

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

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

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