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)
=> \(CM_{HCl}=\dfrac{0,1}{0,7}=0,14M\)
\(CM_{FeCl_2}=\dfrac{0,1}{0,7}=0,14M\)
\(CM_{MgCl_2}=\dfrac{0,2}{0,7}=0,29M\)
