nCaCl2=0.02(mol)

nAgNO3=0.01(mol)

CaCl2+2AgNO3->Ca(NO3)2+2AgCl

Theo pthh nAgNO3=2nCaCl2

Theo bài ra nAgNO3=0.5 nCaCl2

->CaCl2 dư tính theo AgNO3

nAgCl=nAgNO3->nAgCl2=0.01(mol)

mAgCl2=1.435(g)

nCaCl2 phản ứng:0.005(mol)

nCaCl2 dư=0.02-0.005=0.015(mol)->CM=0.015:(0.03+0.07)=0.15M

nCa(NO3)2=0.005(mol)->CM=0.005:(0.03+0.07)=0.05M

$n_{CaCl_2} = \dfrac{2,22}{111} = 0,02(mol) ; n_{AgNO_3} = \dfrac{1,7}{170} = 0,01(mol)$

$CaCl_2 + 2AgNO_3 \to 2AgCl + Ca(NO_3)_2$

Ta thấy : $n_{CaCl_2} : 1 > n_{AgNO_3} : 2$ nên $CaCl_2$ dư

$n_{AgCl} = n_{AgNO_3} = 0,01(mol)$
$m_{AgCl} = 0,01.143,5 = 1,435(gam)$

a, PT: \(CaCl_2+2AgNO_3\rightarrow2AgCl_{\downarrow}+Ca\left(NO_3\right)_2\)

b, Ta có: \(n_{CaCl_2}=\dfrac{2,22}{111}=0,02\left(mol\right)\)

\(n_{AgNO_3}=\dfrac{1,7}{170}=0,01\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,02}{1}>\dfrac{0,01}{2}\), ta được CaCl₂ dư.

Theo PT: \(n_{AgCl}=n_{AgNO_3}=0,01\left(mol\right)\Rightarrow m_{AgCl}=0,01.143,5=1,435\left(g\right)\)

c, \(n_{CaCl_2\left(pư\right)}=\dfrac{1}{2}n_{AgNO_3}=0,005\left(mol\right)\)

\(\Rightarrow n_{CaCl_2\left(dư\right)}=0,015\left(mol\right)\Rightarrow m_{CaCl_2\left(dư\right)}=0,015.111=1,665\left(g\right)\)

Bài 2:

\(a.n_{CaCl_2}=\dfrac{22,2}{111}=0,2\left(mol\right)\\ n_{AgNO_3}=\dfrac{1,7}{170}=0,01\left(mol\right)\\ CaCl_2+2AgNO_3\rightarrow Ca\left(NO_3\right)_2+2AgCl\\ a.Vì:\dfrac{0,2}{1}>\dfrac{0,01}{2}\Rightarrow CaCl_2dư\\b.n_{AgCl}=n_{AgNO_3}=0,01\left(mol\right)\\ \Rightarrow m_{\downarrow}=m_{AgCl}=143,5.0,01=1,435\left(g\right)\)

Bài 1:

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ n_{HCl}=0,1.3=0,3\left(mol\right)\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ a,Vì:\dfrac{0,3}{2}< \dfrac{0,2}{1}\Rightarrow Zndư\\ b.n_{ZnCl_2}=\dfrac{0,3}{2}=0,15\left(mol\right)\\ m_{ZnCl_2}=136.0,15=20,4\left(g\right)\)

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 : 

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

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

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

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