\(PTHH:CaCl_2+2AgNO_3\rightarrow Ca\left(NO_3\right)_2+2AgCl\downarrow\)

\(n_{CaCl_2}=\frac{22,2}{111}=0,2\left(mol\right)\)

Theo PT: \(n_{AgNO_3}=2n_{CaCl_2}=0,4\left(mol\right)\)

\(\Rightarrow m_{AgNO_3}=0,4.170=68\left(g\right)\)

b) Các chất còn lại trong phản ứng là Ca(NO₃)₂, AgCl

\(TheoPT:n_{Ca\left(NO_3\right)_2}=n_{CaCl_2}=0,2\left(mol\right)\)

\(n_{AgCl}=2n_{CaCl_2}=0,4\left(mol\right)\)

\(\Rightarrow m_{Ca\left(NO_3\right)_2}=0,2.164=32,8\left(g\right)\)

\(m_{AgCl}=0,4.143,5=57,4\left(g\right)\)

CaCl₂ + 2AgNO₃ → Ca(NO₃)₂ + 2AgCl

\(n_{CaCl_2}=\frac{22,2}{111}=0,2\left(mol\right)\)

a) Theo PT: \(n_{AgNO_3}=2n_{CaCl_2}=2\times0,2=0,4\left(mol\right)\)

\(\Rightarrow m_{AgNO_3}=0,4\times170=68\left(g\right)\)

b) Theo PT: \(n_{Ca\left(NO_3\right)_2}=n_{CaCl_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{Ca\left(NO_3\right)_2}=0,2\times164=32,8\left(g\right)\)

Theo PT: \(n_{AgCl}=n_{AgNO_3}=0,4\left(mol\right)\)

\(\Rightarrow m_{AgCl}=0,4\times143,5=57,4\left(g\right)\)

Na₂CO₃ + CaCl₂ → 2NaCl + CaCO₃

\(n_{Na_2CO_3}=\frac{10,6}{106}=0,1\left(mol\right)\)

Theo pT: \(n_{CaCO_3}=n_{Na_2CO_3}=0,1\left(mol\right)\)

\(\Rightarrow m_{CaCO_3}=0,1\times100=10\left(g\right)\)

Theo pT: \(n_{NaCl}=2n_{Na_2CO_3}=2\times0,1=0,2\left(mol\right)\)

\(\Rightarrow m_{NaCl}=0,2\times58,5=11,7\left(g\right)\)

\(n_{CaCl_2}=\dfrac{22.2}{111}=0.2\left(mol\right)\)

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

\(0.2....................................................0.4\)

\(m_{AgCl}=0.4\cdot143.5=57.4\left(g\right)\)

n_CaCl2=22,2/111=0,2(mol)

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

TPT:n_AgCl=2.n_CaCl2=2.0,2=0,4(mol)

m_AgCl=0,4.143,5=57,4(g)

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:

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

\(a,n_{CaCl_2}=0,2\cdot0,1=0,02\left(mol\right)\\ n_{AgNO_3}=0,1\cdot0,1=0,01\left(mol\right)\\ PTHH:CaCl_2+2AgNO_3\rightarrow2AgCl\downarrow+Ca\left(NO_3\right)_2\\ \text{Vì }\dfrac{n_{CaCl_2}}{1}>\dfrac{n_{AgNO_3}}{2}\Rightarrow CaCl_2\text{ dư}\\ \Rightarrow n_{AgCl}=0,01\left(mol\right)\\ \Rightarrow m_{AgCl}=0,01\cdot143,5=1,435\left(g\right)\\ b,n_{Ca\left(NO_3\right)_2}=\dfrac{1}{2}n_{AgCl}=0,005\left(mol\right)\\ \Rightarrow C_{M_{Ca\left(NO_3\right)_2}}=\dfrac{0,005}{0,1+0,1}=0,025M\)

Mọi người ơi giúp mik vs ạ . cảm ơn mn rất nhiều.

a)

Gọi : \(\left\{{}\begin{matrix}n_{NaCl}=a\left(mol\right)\\n_{KI}=b\left(mol\right)\end{matrix}\right.\)

NaCl + AgNO₃ → AgCl + NaNO₃

a..............a...............a..............................(mol)

KI + AgNO₃→ AgI + KNO₃

b.......b..............b..................................(mol)

Ta có :

\(n_{AgNO_3} = a + b = 0,25.2 = 0,5(mol)\)

\(m_{kết\ tủa} = 143,5a + 235b = 103,775\)(gam)

Suy ra : a = 0,15 ; b = 0,35

Vậy :

\(C_{M_{NaCl}} = \dfrac{0,15}{0,4} = 0,375M\\ C_{M_{KI}} = \dfrac{0,35}{0,4} = 0,875M\)

b)

Sau phản ứng, dung dịch gồm : \(\left\{{}\begin{matrix}NaNO_3:0,15\left(mol\right)\\KNO_3:0,35\left(mol\right)\end{matrix}\right.\)

Suy ra : 

\(m_{NaNO_3} = 0,15.85 = 12,75(gam)\\ m_{KNO_3} = 0,35.101 = 35,35(gam)\)

a) Đặt nAl=a(mol) ; nFe=b(mol) (a,b>0)

nHCl= (365.12%)/36,5=1,2(mol)

PTHH: 2Al + 6 HCl -> 2AlCl3 +3 H2

a________3a_________2a____1,5a(mol)

Fe + 2 HCl -> FeCl2 + H2

b_____2b____b____b(mol)

Ta có hpt:

\(\left\{{}\begin{matrix}27a+56b=22,2\\3a+2b=1,2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,3\end{matrix}\right.\)

b) => %mAl= [(0,2.27)/22,2].100=24,324%

=>%mFe= 75,676%

c) mFeCl2=127. 0,3=38,1(g)

mAlCl3= 133,5. 0,2= 26,7(g)

mddsau= 22,2+365 - 1,2.2=384,8(g)

=>C%ddFeCl2= (38,1/384,8).100=9,901%

C%ddAlCl3= (26,7/384,8).100=6,939%

PTHH: \(2AgNO_3+CaCl_2\rightarrow Ca\left(NO_3\right)_2+2AgCl\downarrow\)

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

\(\Rightarrow\left\{{}\begin{matrix}n_{AgCl}=0,01\left(mol\right)\\n_{CaCl_2}=n_{Ca\left(NO_3\right)_2}=0,005\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{CaCl_2}=0,005\cdot111=0,555\left(g\right)\\m_{AgCl}=0,01\cdot143,5=1,435\left(g\right)\\C_{M_{Ca\left(NO_3\right)_2}}=\dfrac{0,005}{0,07+0,03}=0,05\left(M\right)\end{matrix}\right.\)

\(a,PTHH:R+2AgNO_3\to R(NO_3)_2+2Ag\\ \Rightarrow n_{R}=n_{R(NO_3)_2}\\ \Rightarrow \dfrac{2,8}{M_R}=\dfrac{9}{M_R+124}\\ \Rightarrow M_R=56(g/mol)\)

Vậy R là sắt (Fe)

\(b,n_{R}=\dfrac{2,8}{56}=0,05(mol)\\ \Rightarrow n_{AgNO_3}=0,1(mol)\\ \Rightarrow m_{dd_{AgNO_3}}=\dfrac{0,1.170}{5\%}=340(g)\\ c,n_{Fe(NO_3)_2}=n_{Fe}=0,05(mol);n_{Ag}=0,1(mol)\\ \Rightarrow C\%_{Fe(NO_3)_2}=\dfrac{0,05.180}{2,8+340-0,1.108}.100\%=2,71\%\)

-nNa₂CO₃= m/M = 10,6/106 = 0,1 (mol)

-PT:Na₂CO₃+CaCl₂->CaCO₃+2NaCl

____0,1____________0,1______0,2

-mCaCO₃= n.M = 0,1.100 = 10 (g)

-mNaCl= n.M = 0,2.58,5 = 11,7 (g)

\(n_{Na_2CO_3}=\dfrac{10.6}{106}=0.1\left(mol\right)\\ PTHH:CaCl_2+Na_2CO_3\rightarrow CaCO_3+2NaCl\\ n_{CaCO_3}=n_{Na_2CO_3}=0,1\left(mol\right)\\ \Rightarrow m_{CaCO_3}=0,1.100=10g\)

$n_{NaCl}=n_{Na_2CO_3}=0,1.2=0,2(mol)$

`=>` $m_{NaCl}=0.2.58,5=11,7g$