\(a.n_{CaCl_2}=0,1\left(mol\right)\\ CaCl_2+2AgNO_3\rightarrow Ca\left(NO_3\right)_2+2AgCl\\ n_{AgCl}=2n_{CaCl_2}=0,2\left(mol\right)\\ \Rightarrow m_{AgCl}=0,2.143,5=28,7\left(g\right)\\ b.n_{Ca\left(NO_3\right)_2}=n_{CaCl_2}=0,1\left(mol\right)\\ \Rightarrow CM_{CaCl_2}=\dfrac{0,1}{0,1}=1M\)

\(n_{BaCl_2}=0.2\cdot0.5=0.1\left(mol\right)\)

\(BaCl_2+K_2SO_4\rightarrow BaSO_4+2KCl\)

\(0.1.............0.1.........................0.2\)

\(V_{dd_{K_2SO_4}}=\dfrac{0.1}{1}=0.1\left(l\right)\)

\(V_{dd}=0.2+0.1=0.3\left(l\right)\)

\(C_{M_{KCl}}=\dfrac{0.2}{0.3}=0.67\left(M\right)\)

Đổi 200ml = 0,2 lít

Ta có: \(n_{BaCl_2}=0,5.0,2=0,1\left(mol\right)\)

a. PTHH: \(BaCl_2+K_2SO_4--->BaSO_4\downarrow+2KCl\)

Theo PT: \(n_{K_2SO_4}=n_{BaCl_2}=0,1\left(mol\right)\)

\(\Rightarrow V_{dd_{K_2SO_4}}=\dfrac{0,1}{1}=0,1\left(lít\right)\)

b. Theo PT: \(n_{KCl}=2.n_{BaCl_2}=2.0,1=0,2\left(mol\right)\)

Ta có: \(V_{dd_{KCl}}=V_{dd_{BaCl_2}}=0,1\left(lít\right)\)

\(\Rightarrow C_{M_{KCl}}=\dfrac{0,2}{0,1}=2M\)

\(n_{NaCl}=\dfrac{3,51}{58,5}=0,06\left(mol\right)\)

a) Pt : \(NaCl+AgNO_3\rightarrow NaNO_3+AgCl|\)

              1               1                1             1

           0,06            0,06          0,06         0,06

a) \(n_{AgCl}=\dfrac{0,06.1}{1}=0,06\left(mol\right)\)

⇒ \(m_{AgCl}=0,06.143,5=8,61\left(g\right)\)

b) \(n_{AgNO3}=\dfrac{0,06.1}{1}=0,06\left(mol\right)\)

\(V_{ddAgNO3}=\dfrac{0,06}{0,2}=0,3\left(l\right)\)

c) \(n_{NaNO3}=\dfrac{0,06.1}{1}=0,06\left(mol\right)\)

\(C_{M_{NaNO3}}=\dfrac{0,06}{0,3}=0,2\left(M\right)\)

 Chúc bạn học tốt

\(a,n_{NaCl}=\dfrac{3,51}{58,5}=0,06(mol)\\ PTHH:NaCl+AgNO_3\to AgCl\downarrow+NaNO_3\\ \Rightarrow n_{AgCl}=0,06(mol)\\ \Rightarrow m_{AgCl}=0,06.143,5=8,61(g)\\ b,n_{AgNO_3}=0,06(mol)\\ \Rightarrow V_{dd_{AgNO_3}}=\dfrac{0,06}{0,2}=0,3(l)\\ c,n_{NaNO_3}=0,06(mol);V_{dd_{NaNO_3}}=V_{dd(\text {phản ứng})}=0,3(l)\\ \Rightarrow C_{M_{NaNO_3}}=\dfrac{0,06}{0,3}=0,2M\)

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

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

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

Xem lại đề

\(FeCl_x+xAgNO_3\rightarrow Fe\left(NO_3\right)_x+xAgCl\downarrow\)

\(n_{AgCl}=\frac{21,525}{143,5}=0,15\left(mol\right)\)

Theo PT \(n_{FeCl_x}=\frac{0,15}{x}\left(mol\right)\)

Ta có: \(n_{FeCl_x}=\frac{190,5.10\%}{56+35,5x}=\frac{0,15}{x}\left(mol\right)\)

=> x=

a) Zn+Cl2---->ZnCl2

Ca+Cl2---->CaCl2

b) Gọi n Zn=x------>m Zn=65x

n Ca=y--------->m Ca=40y

--->65x+40y=1,45(1)

n Cl2=0,672/22,4=0,03(mol)

--->x+y=0,03(2)

Từ 1 và 2 ta có hpt

\(\left\{{}\begin{matrix}65x+40y=1,45\\x+y=0,03\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,01\\y=0,02\end{matrix}\right.\)

%m Zn=0,01.65/1,45.100%=44,83%

%m Ca=100-44,83=55,17%

c)ZnCl2+2AgNO3--->2AgCl+Zn(NO3)2

0,1-------------------->0,2(mol)

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

0,2----------------------->0,4(mol)

Tổng n AgCl=0,6(mol)

m AgCl=0,6.143,5=86,1(g)

Chúc bạn học tốt

$a\big)$

$Zn+2HCl\to ZnCl_2+H_2$

$CuO+H_2\xrightarrow{t^o}Cu+H_2O$

$b\big)$

$n_{Zn}=\dfrac{10,4}{65}=0,16(mol)$

Theo PT: $n_{Cu}=n_{Zn}=0,16(mol)$

$\to m_{Cu}=0,16.64=10,24(g)$

a, Ta có: \(\%NaCl=\frac{m_{ct}}{m_{dd}}\cdot100\)

\(\Leftrightarrow9=\frac{m_{ct}}{300}\cdot100\)

\(\rightarrow mct=27\left(g\right)\)

\(\rightarrow n_{NaCl}=\frac{27}{50}=0.54mol\)

PTHH: \(NaCl+AgNO_3\rightarrow NaNO_3+AgCl\downarrow\)

............0.54mol\(\rightarrow\) 0.54mol \(\rightarrow\) 0.54mol \(\rightarrow\) 0.54mol

\(m_{AgCl}=0.54\cdot143.5=77.49\left(g\right)\)

c, \(m_{AgNO3}=0.54\cdot168=90.72\left(g\right)\)

từ 9% trong 300 g dd NaCl tính ra khối lượng rồi tính số mol của NaCl

Sau đó tính theo PTHH và giải ra