Chọn A

mình cần cách tính cụ thể hơn

cách tính kiểu gì

tham khảo 

a,nA=\(\dfrac{18,25}{36,5}\)=0,5(mol)

nB=\(\dfrac{10,95}{36,5}\)=0,3(mol)

→nC=0,3+0,5=0,8(mol)

→CM(C)=\(\dfrac{0,8}{2}\)=0,4M

b,CM(A)=\(\dfrac{0,5}{V1}\)

CM(B)=\(\dfrac{0,3}{V2}\)

→\(\dfrac{0,5}{V1}\)=\(\dfrac{0,3}{V2}\)=0,8

=>V1=0,625  l

=>V2=0,375 l 

=>CmV1=\(\dfrac{0,5}{0,625}\)=0,8M

=>CmV2=\(\dfrac{0,3}{0,375}\)=0,8M

\(a,n_A=\dfrac{18,25}{36,5}=0,5\left(mol\right)\\ n_B=\dfrac{10,95}{36,5}=0,3\left(mol\right)\)

\(\rightarrow n_C=0,3+0,5=0,8\left(mol\right)\\ \rightarrow C_{M\left(C\right)}=\dfrac{0,8}{2}=0,4M\)

\(b,C_{M\left(A\right)}=\dfrac{0,5}{V_1}\\ C_{M\left(B\right)}=\dfrac{0,3}{V_2}\\ \rightarrow\dfrac{0,5}{V_1}:\dfrac{0,3}{V_2}=0,8\\ \rightarrow\dfrac{0,5}{V_1}=\dfrac{0,24}{V_2}=\dfrac{0,5+0,24}{V_1+V_2}=\dfrac{0,74}{2}=0,37\\ \rightarrow\left\{{}\begin{matrix}V_1=\dfrac{0,5}{0,34}=1,4\left(l\right)\\V_2=\dfrac{0,24}{0,34}=0.6\left(l\right)\end{matrix}\right.\\ \rightarrow\left\{{}\begin{matrix}C_{M\left(A\right)}=\dfrac{0,5}{1,4}=0,36M\\C_{M\left(B\right)}=\dfrac{0,5}{0,6}=0,83M\end{matrix}\right.\)

a) \(n_{HCl\left(A\right)}=\dfrac{7,3}{36,5}=0,2\left(mol\right)\)

\(n_{HCl\left(B\right)}=\dfrac{58,4}{36,5}=1,6\left(mol\right)\)

=> \(n_{HCl\left(C\right)}=0,2+1,6=1,8\left(mol\right)\)

=> \(C_{M\left(C\right)}=\dfrac{1,8}{3}=0,6M\)

b) 
\(C_{M\left(A\right)}=\dfrac{0,2}{V_1}M\)

\(C_{M\left(B\right)}=\dfrac{1,6}{V_2}M\)

=> \(\dfrac{1,6}{V_2}-\dfrac{0,2}{V_1}=0,6\)

=> \(\dfrac{1,6}{3-V_1}-\dfrac{0,2}{V_1}=0,6\)

=> \(1,6.V_1-0,2\left(3-V_1\right)=0,6.V_1.\left(3-V_1\right)\)

=> \(1,6.V_1-0,6+0,2.V_1=1,8.V_1-0,6.V_1^2\)

=> \(0,6.V_1^2=0,6\)

=> V₁ = 1 (l)

=> V₂ = 2 (l)

\(C_{M\left(A\right)}=\dfrac{0,2}{1}=0,2M\)

\(C_{M\left(B\right)}=\dfrac{1,6}{2}=0,8M\)

a)Theo đề bài ta có
mct=mHCl=(mdd.C%)/100%=(150.2,65%)/100%=3,975 g
-> nHCl=3,975/36,5=0,1mol
-> Nồng độ mol của dung dịch thu được là
C%=n/V=0,1/2=0,05 M

Ta biết là Dnước = 1

Gọi thể tích dd H2SO4 (D=1,84) và nước cần dùng là x, y (l) thì ta có

{x+y=101000x.1,84+y=10000.1,28

⇔{x+y=101840x+y=12800

⇔{x=6,95y=3,05

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15mol\)

2Al+2NaOH+2H₂O\(\rightarrow\)2NaAlO₂+3H₂

\(n_{Al}=\dfrac{2}{3}n_{H_2}=\dfrac{2}{3}.0,15=0,1mol\)\(\rightarrow\)\(m_{Al}=2,7g\)

\(m_{Mg,Fe}=14,7-2,7=12g\)

\(n_{Mg}=xmol\);\(n_{Fe}=ymol\)

Mg+2HCl\(\rightarrow\)MgCl₂+H₂

Fe+2HCl\(\rightarrow\)FeCl₂+H₂

- Ta có hệ phương trình:

\(\left\{{}\begin{matrix}24x+56y=12\\x+y=\dfrac{10,08}{22,4}=0,45\end{matrix}\right.\)

Giải ra x=0,4125 và y=0,0375

\(\%Al=\dfrac{2,7}{14,7}.100\approx18,4\%\)

\(\%Mg=\dfrac{0,4125.24}{14,7}.100\approx67,35\%\)

\(\%Fe=100\%-18,4\%-67,35\%=14,25\%\)

- Tóm tắt PTHH:

MgCl₂\(\rightarrow\)Mg(OH)₂\(\rightarrow\)MgO

FeCl₂\(\rightarrow\)Fe(OH)₂\(\rightarrow\)Fe(OH)₃\(\rightarrow\)Fe₂O₃

\(m=m_{MgO}+m_{Fe_2O_3}=0,4125.40+\dfrac{1}{2}.0,0375.160=19,5g\)