Mg +2HCl\(\rightarrow\)MgCl2+ H2

MgO+2HCl\(\rightarrow\)MgCl2+H2O

ta có nH2 =\(\frac{2,24}{22,4}\)=0,1 mol

Theo pthh 1 :nH2=nMg =0,1 mol

\(\rightarrow\)mMg=0,1.24 =2,4 g

\(\rightarrow\)mMgO =6-2,4=3,6g

nMgO =\(\frac{3,6}{40}\)=0,09 mol

%mMgO =\(\frac{3,6}{6}.100\%\)=60%

Ta co nHCl =2nMg =0,2 mol

nHCl =2nMgO =0,18 mol

mHCl=(0,18+0,2).36,5=13,87 g

VHCl=\(\frac{13,87}{1}.1\)=12,6ml

Bài 1: Gọi kl hóa trị II là A; kl hóa trị III là B

\(A+2HCl\rightarrow ACl_2+H_2\)

\(2B+6HCl\rightarrow2BCl_3+3H_2\)

\(n_{HCl}=0,5.0,02=0,01mol\)

Theo ĐLBTKL: m_{muối khan} = \(4,6+0,01.36,5-0,005.2=4,955g\)

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

\(MgO+2HCl\rightarrow MgCl_2+H_2o\)

\(n_{H_2}=\dfrac{1,12}{22,4}=0,05mol=n_{Mg}\)

\(\Rightarrow m_{Mg}=1,2g\Rightarrow m_{MgO}=2g\)

\(\Rightarrow\%m_{Mg}=\dfrac{1,2.100}{3,2}=37,5\%\)

\(\Rightarrow\%m_{MgO}=62,5\%\)

\(m_{ddsaupư}=3,2+246,9-0,05.2=250g\)

\(C\%ddMgCl_2=\dfrac{0,05.96}{250}.100\%=1,92\%\)

\(\text{nCO2= 4,48:22,4=0,2 mol}\)

MgO + 2HCl -> MgCl2+ H2O

MgCO3 +2 HCl -> MgCl2+ H2O + CO2

mol 0,2....<- 0,4.....<-0,2 .................<-0,2

\(\Rightarrow\text{ mMgCO3=0,2.84=16,8(g) }\)

\(\Rightarrow\text{mMgO=20,8-16,8=4g}\)

\(\Rightarrow\text{%MgCO3= 16,8:20,8.100%=80,77%}\)

\(\Rightarrow\text{ %MgO=100-80,77=19,23%}\)

\(\text{b, nMgO=4÷ 0,1(mol)}\)

Theo PTHH 1: n HCl= 2n MgO= 0,1.2= 0,2

\(\text{nHCl = 2nMgCO3= 0,2.2= 0,4}\)

\(\Rightarrow\text{nHCl=0,2+0,4=0,6}\)

\(\Rightarrow\text{mHCl=0,6. 36,5=21,9g}\)

\(\Rightarrow\text{mddHCl= 21,9.100:7,3=300g}\)

\(\Rightarrow\text{ V= 300: 1,05=285,71g}\)

MgO+2HCl---->MgCl2+H2

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

n\(_{CO2}=\frac{4,48}{22,4}=0,2\left(mol\right)\)

Theo pthh2

n\(_{MgCO3}=n_{CO2}=0,2\left(mol\right)\)

%m MgCO3=\(\frac{0,2.84}{20,8}.100\%=80,77\%\)

%m MgO=100-80,77=19,23%

b) mol HCl ở PT1 là n\(_{HCl}=2n_{MgO}=\frac{4}{40}.2=0,2\left(mol\right)\)

mol HCl ở pt2 là n\(_{HCl}=2n_{CO2}=0,4\left(mol\right)\)

Tổng mol HCl =0,2+0,4=0,6(mol)

m\(_{ddHCl}=\frac{0,6.36,5.100}{7,3}=300\left(g\right)\)

V HCl=300/1,05=285,71(l)

c) Theo pthh1

n\(_{MgCl2}=n_{MgO}=0,1\left(mol\right)\)

m dd sau pư=300+20,8-8,8=312(g)

Theo pthh2

n\(_{MgCl2}=n_{CO2}=0,2\left(mol\right)\)

C% MgCl2 =\(\frac{0,3.95}{312}.100\%=9,13\%\)

\(a.PTHH:\)

\(Mg+2HCl--->MgCl_2+H_2\uparrow\left(1\right)\)

\(MgO+2HCl--->MgCl_2+H_2O\left(2\right)\)

b. ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

Theo PT₍₁₎: \(n_{Mg}=n_{H_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{MgO}=12-0,2.24=7,2\left(g\right)\)

\(\Rightarrow\%_{MgO}=\dfrac{7,2}{12}.100\%=60\%\)

c. Ta có: \(n_{hh}=0,2+\dfrac{7,2}{40}=0,38\left(mol\right)\)

Theo PT_(1,2): \(n_{HCl}=2.n_{hh}=2.0,38=0,76\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,76.36,5=27,74\left(g\right)\)

\(\Rightarrow m_{dd_{HCl}}=138,7\left(g\right)\)

\(\Rightarrow V_{dd_{HCl}}=126\left(ml\right)\)

Ta có:

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

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

\(MgO+2HCl\rightarrow MgCl_2+H_2O\)

\(\Rightarrow n_{Mg}=n_{H2}=0,1\left(mol\right)\Rightarrow m_{Mg}=0,1.24=2,4\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\frac{2,4}{12}.100\%=20\%\\\%m_{MgO}=100\%-20\%=80\%\end{matrix}\right.\)

\(n_{MgO}=6-2,4=3,6\left(g\right)\)

\(\Rightarrow n_{MgO}=\frac{3,6}{40}=0,09\left(mol\right)\)

\(n_{HCl}=0,38\left(mol\right)\Rightarrow m_{HCl}=0,38.36,5=13,87\left(g\right)\)

\(\Rightarrow V_{HCl}=\frac{13,87}{1,1}=12,6\left(l\right)\)

Mà dd HCl chỉ bằng 20% nên:

\(\Rightarrow V_{HCl}=\frac{12,6.20}{100}=2,512\left(ml\right)\)