30ml = 0,03l

70ml = 0,07l

\(n_{CaCl2}=0,5.0,03=0,015\left(mol\right)\)

\(n_{AgNO3}=1.0,07=0,07\left(mol\right)\)

Pt : \(CaCl_2+2AgNO_3\rightarrow Ca\left(NO_3\right)_2+2AgCl|\)

           1              2                 1                    2

      0,015         0,07             0,015             0,03

Lập tỉ số so sánh : \(\dfrac{0,015}{1}< \dfrac{0,07}{2}\)

                 ⇒ CaCl₂ phản ứng hết , AgNO₃ dư

                ⇒ Tính toán dựa vào số mol của CaCl₂ 

\(n_{AgCl}=\dfrac{0,015.2}{1}=0,03\left(mol\right)\)

⇒ \(m_{AgCl}=0,03.143,5=4,305\left(g\right)\)

\(n_{Ca\left(NO3\right)2}=\dfrac{0,03.1}{2}=0,015\left(mol\right)\)

\(n_{AgNO3\left(dư\right)}=0,07-\left(0,015.2\right)=0,04\left(mol\right)\)

Sau phản ứng : 

\(V_{dd}=0,03+0,07=0,1\left(l\right)\)

\(C_{M_{Ca\left(NO3\right)2}}=\dfrac{0,015}{0,1}=0,15\left(M\right)\)

\(C_{M_{AgNO3\left(dư\right)}}=\dfrac{0,04}{0,1}=0,4\left(M\right)\)

 Chúc bạn học tốt

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,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

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

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

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