\(n_{Fe}=\dfrac{11.2}{56}=0.2\left(mol\right)\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(0.2...................0.2..........0.2\)

\(V_{H_2}=0.2\cdot22.4=4.48\left(l\right)\)

\(m_{FeCl_2}=0.2\cdot127=25.4\left(g\right)\)

\(m_{\text{dung dịch sau phản ứng}}=11.2+200-0.2\cdot2=210.8\left(g\right)\)

\(C\%_{FeCl_2}=\dfrac{25.4}{210.8}\cdot100\%=12.05\%\)

Giúp với mai mình thi rồi!

\(n_{Fe}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

PT :

\(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)

0,1     0,2           0,1         0,1 

\(a,V_{H_2}=0,1.22,4=2,24\left(l\right)\)

\(b,m_{HCl}=0,2.36,5=7,3\left(g\right)\)

\(C\%=\dfrac{7,3}{73}.100\%=10\%\)

\(c,m_{ddZnCl_2}=6,5+73-\left(0,1.2\right)79,3\left(g\right)\)

\(m_{ZnCl_2}=0,1.136=13,6\left(g\right)\)

\(C\%_{ZnCl_2}=\dfrac{13,6}{79,3}.100\%=17,15\%\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)

0,1     2

Vì 0,1/1<2/2

nên tính theo Zn

=>\(n_{H_2}=n_{Zn}=0.1\left(mol\right)\) và n_HCl=0,2(mol)

\(V=0.1\cdot22.4=2.24\left(lít\right)\)

\(C\%\left(muối\right)=\dfrac{0.1\cdot136}{6.5+73-0.2}\simeq17,15\%\)

a) PTHH: Mg +2 HCl -> MgCl2 + H2

 nH2=11,2/22,4=0,5(mol)

=> nMg=nMgCl2=nH2=0,5(mol)

m=mMg=0,5.24=12(g)

b) mMgCl2=0,5.95=47,5(g)

mddMgCl2= mMg + mddHCl - mH2= 12+100-0,5.2= 111(g)

=>C%ddMgCl2= (47,5/111).100=42,793%

\(n_{MnO_2}=\dfrac{69,6}{87}=0,8\left(mol\right)\)

n_KOH = 0,5.4 = 2(mol)

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

             0,8------------------------>0,8

           2KOH + Cl₂ --> KCl + KClO + H₂O

Xét tỉ lệ \(\dfrac{2}{2}>\dfrac{0,8}{1}\) => KOH dư, Cl₂ hết

2KOH + Cl₂ --> KCl + KClO + H₂O

 1,6<--0,8---->0,8---->0,8

=> \(\left\{{}\begin{matrix}n_{KOH\left(dư\right)}=2-1,6=0,4\left(mol\right)\\n_{KCl}=0,8\left(mol\right)\\n_{KClO}=0,8\left(mol\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}C_{M\left(KOH\right)}=\dfrac{0,4}{0,5}=0,8M\\C_{M\left(KCl\right)}=\dfrac{0,8}{0,5}=1,6M\\C_{M\left(KClO\right)}=\dfrac{0,8}{0,5}=1,6M\end{matrix}\right.\)

ta có 200cm3=0,2 lítn hcl=2*0,2=0,4 molgọi số mol của caco3 là a,na2co3 là bcaco3 + 2hcl -> cacl2 + co2 + h2oa(mol)---2a(mol)--a-------a--------ana2co3 + 2hcl -> 2nacl + co2 + h2ob(mol)---2b(mol)---2b-------b------bta có100a+106b=20,62a+2b=0,4=> a=b=0,1 mol=> m caco3=10g; m na2co3=10,6 g 

PTHH: \(Na_2CO_3+2HCl\rightarrow2NaCl+H_2O+CO_2\uparrow\)  (1)

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

a) Ta có: \(\Sigma n_{HCl}=0,2\cdot2=0,4\left(mol\right)\)

Gọi số mol của Na₂CO₃ là \(a\) \(\Rightarrow n_{HCl\left(1\right)}=2a\left(mol\right)\)

Gọi số mol của CaCO₃ là \(b\) \(\Rightarrow n_{HCl\left(2\right)}=2b\left(mol\right)\)

Ta lập được hệ phương trình:

\(\left\{{}\begin{matrix}2a+2b=0,4\\106a+100b=20,6\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{CaCO_3}=0,1\cdot100=10\left(g\right)\\m_{Na_2CO_3}=10,6\left(g\right)\end{matrix}\right.\)

b) Theo PTHH: \(\left\{{}\begin{matrix}n_{NaCl}=2n_{Na_2CO_3}=0,2mol\\n_{CaCl_2}=n_{CaCO_3}=0,1mol\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{NaCl}=0,2\cdot58,5=11,7\left(g\right)\\m_{CaCl_2}=0,1\cdot111=11,1\left(g\right)\end{matrix}\right.\)

Mặt khác: \(\left\{{}\begin{matrix}m_{CO_2}=0,2\cdot44=8,8\left(g\right)\\m_{ddHCl}=200\cdot1,2=240\left(g\right)\end{matrix}\right.\)

\(\Rightarrow m_{dd\left(saup/ư\right)}=m_{hh}+m_{ddHCl}-m_{CO_2}=251,8\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{NaCl}=\dfrac{11,7}{251,8}\cdot100\%\approx4,65\%\\C\%_{CaCl_2}=\dfrac{11,1}{251,8}\cdot100\%\approx4,41\%\end{matrix}\right.\)

m_Na2CO3 = \(\frac{md.C\%}{100\%}\)= \(\frac{100.10,6\%}{100\%}\)= 10,6 (g)

n_Na2CO3 = \(\frac{m}{M}\)= \(\frac{10,6}{106}\)= 0,1 (mol)

m_H2SO4 = \(\frac{md.C\%}{100\%}\)= \(\frac{200.9,8\%}{100\%}\)= 19,6 (g)

n_H2SO4 = \(\frac{m}{M}\)= \(\frac{19,6}{98}\)= 0,2 (mol)

Khi cho Na₂CO₃ tác dụng với H₂SO₄, ta có PTHH:

Na₂CO₃ + H₂SO₄  \(\rightarrow\)Na₂SO₄ + CO₂\(\uparrow\)+ H₂O

0,1          : 0,2

Xét tỉ lệ: \(\frac{0,1}{1}\)< \(\frac{0,2}{1}\)=> H₂SO₄ dư, dưa vào n_Na2CO3 để tính

Na₂CO₃ + H₂SO₄ \(\rightarrow\) Na₂SO₄ + CO₂\(\uparrow\)+ H₂O

0,1 \(\rightarrow\)0,1              : 0,1            : 0,1        : 0,1               (mol)

a. V_CO2 = n.22,4 = 0,1.22,4 = 2,24 (l)

b. Các chất sau phản ứng gồm Na₂SO₄ và H₂SO₄ dư

m_dsau = m_dNa2CO3 + m_dH2SO4 - m_CO2 = 100 + 200 - 0,1.44 = 295,6 (g)

C%_Na2CO3 = \(\frac{mt}{md}\). 100% = \(\frac{0,1.106}{295,6}\). 100% \(\approx\)3,6 %

C%_H2SO4 = \(\frac{mt}{md}\). 100% = \(\frac{\left(0,2-0,1\right)98}{295,6}\). 100% \(\approx\)3,31%

\(a)n_{Fe}=\dfrac{25,2}{56}=0,45mol\\ Fe+2HCl\rightarrow FeCl_2+H_2\)

0,45       0,9          0,45       0,45

\(V_{H_2\left(đktc\right)}=0,45.22,4=10,08l\\ b)C_{M\left(HCl\right)}=\dfrac{0,9}{0,2}=4,5M\\ c)C_{M\left(FeCl_2\right)}=\dfrac{0,45}{0,2}=2,25M\)

Gọi \(n_{Fe}=x\left(mol\right)\)\(;n_{Zn}=y\left(mol\right)\)

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3mol\)

Ta có:  \(\left\{{}\begin{matrix}56x+65y=18,6\\2x+2y=2n_{H_2}=0,6\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)

\(\%m_{Fe}=\dfrac{0,1\cdot56}{18,6}\cdot100\%=30,11\%\)

\(\%m_{Zn}=100\%-30,11\%=69,89\%\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

0,1     0,2

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

0,2     0,4

\(n_{HCl}=0,2+0,4=0,6mol\)

\(C_M=\dfrac{n}{V}=\dfrac{0,6}{0,2}=3M\)