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)

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

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

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

Đề câu 2 sao khỏi làm đi

Bài này ta phải nhìn kỹ vào PTHH và biết vận dụng ĐLBTKL :)

\(n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH:

\(MgCO_3+2HCl\rightarrow MgCl_2+CO_2\uparrow+H_2O\left(1\right)\\ CaCO_3+2HCl\rightarrow CaCl_2+CO_2\uparrow+H_2O\left(2\right)\\ Na_2CO_3+2HCl\rightarrow2NaCl+CO_2\uparrow+H_2O\left(3\right)\)

Theo pthh (1, 2, 3): \(\left\{{}\begin{matrix}n_{H_2O}=n_{CO_2}=0,2\left(mol\right)\\n_{HCl}=2n_{CO_2}=2.0,2=0,4\left(mol\right)\end{matrix}\right.\)

Áp dụng ĐLBTKL: 

\(m_{muối.cacbonat}+m_{HCl}=m_{muối.clorua}+m_{CO_2}+m_{H_2O}\)

=> x = m_{muối clorua} = 18,7 + 0,4.36,5 - 0,2.44 - 0,2.18 = 20,9 (g)

\(Đặt:nMg=a\left(mol\right),n_{MgCO_3}=b\left(mol\right)\)

\(n_{khí}=\dfrac{6.72}{22.4}=0.3\left(mol\right)\)

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

\(a................a.......a\)

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

\(b....................b........b\)

\(m_X=24a+84b=13.2\left(g\right)\left(1\right)\)

\(n_{khí}=a+b=0.3\left(mol\right)\left(2\right)\)

\(\left(1\right),\left(2\right):a=0.2,b=0.1\)

\(m_{Mg}=0.2\cdot24=4.8\left(g\right),m_{MgCO_3}=0.1\cdot84=8.4\left(g\right)\)

\(m_{ddB}=13.2+200-0.2\cdot2-0.1\cdot44=208.4\left(g\right)\)

\(m_{MgCl_2}=0.3\cdot95=28.5\left(g\right)\)

\(C\%MgCl_2=\dfrac{28.5}{208.4}\cdot100\%=13.67\%\)

- Gọi x,y lần lượt là khối lượng của CaCO3 và MgCO trong hh (x,y >0) (g)

nCaCO3= x(mol)=> mCaCO3= 100x(g)

nMgCO3= y(mol) => mMgCO3= 84y(g)

=> mCaCO3+ mMgCO3= mhh

<=> 100x+ 84y= 2,84 (a)

PTHH: (1) CaCO3 + 2 HCl -> CaCl2 + H2O + CO2

(2) MgCO3 + 2 HCl -> MgCl2 + H2O + CO2

Ta có: nCO2(tổng)= (672:1000):22,4= 0,03(mol)

nCO2(1)= nCaCO3= x(mol); nCO2(2)= nMgCO3= y(mol)

=> nCO2(1) + nCO2(2)= nCO2(tổng)

<=> x+y= 0,03 (b)

Từ (a),(b) ta có hpt:

\(\left\{{}\begin{matrix}100x+84y=2,84\\x+y=0,03\end{matrix}\right.\)

Giaỉ hpt: x= 0,02 ; y= 0,01

=> mCaCO3= 100x= 100. 0,02= 2(g)

=> %mCaCO3= (2/ 2,84).100 \(\approx\) 70,423%

=> %mMgCO3 \(\approx\)100% - 70,423%= 29,577%

Đặt \(\left\{{}\begin{matrix}n_{CaCO_3}=x\\n_{MgCO_3}=y\end{matrix}\right.\) ( mol )

\(\Rightarrow m_{hh}=100x+84y=14,2\left(g\right)\)   (1)

\(CaCO_3\rightarrow\left(t^o\right)CaO+CO_2\)

    x                      x          x        ( mol )

\(MgCO_3\rightarrow\left(t^o\right)MgO+CO_2\)

    y                       y            y         ( mol )

\(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3\downarrow+H_2O\)

                        0,15     0,15                 ( mol )

\(n_{CaCO_3}=\dfrac{15}{100}=0,15\left(mol\right)\)

\(\Rightarrow n_{CO_2}=x+y=0,15\left(mol\right)\)          (2)

\(\left(1\right);\left(2\right)\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,05\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\%m_{CaCO_3}=\dfrac{0,1.100}{14,2}.100=70,42\%\\\%m_{MgCO_3}=100-70,42=29,58\%\end{matrix}\right.\)

\(m_{rắn}=m_{CaO}+m_{MgO}=0,1.56+0,05.40=7,6\left(g\right)\)

\(V_{CO_2}=0,15.22,4=3,36\left(l\right)\)