a. PTHH: AgNO₃ + HCl ---> AgCl↓ + HNO₃

b. Ta có: \(n_{AgNO_3}=\dfrac{42,5}{170}=0,25\left(mol\right)\)

Theo PT: \(n_{AgCl}=n_{AgNO_3}=0,25\left(mol\right)\)

=> \(m_{AgCl}=0,25.143,5=35,875\left(g\right)\)

c. Theo PT: \(n_{HCl}=n_{AgCl}=0,25\left(mol\right)\)

Đổi 100ml = 0,1 lít

=> \(C_{M_{HCl}}=\dfrac{0,25}{0,1}=2,5M\)

Của dd axit nha

\(n_{AgNO3}=0,2\left(mol\right)\)

\(n_{CaCl2}=0,15\left(mol\right)\)

PT: 2AgNO3 + CaCl2\(\rightarrow\) 2AgCl\(\downarrow\)+ Ca(NO3)2

0,2 0,1 (mol)

\(\Rightarrow\)\(n_{CaCl2pư}=0,1\left(mol\right)\)

\(\Rightarrow n_{CaCl2}\)dư=0,05(mol)

\(\Rightarrow C_{MCaCl2}\)dư =\(\dfrac{0,05}{0,2+0,1}\)=\(\)0,17M

\(\Rightarrow n_{Ca\left(NO3\right)2}=0,1\left(mol\right)\)

\(\Rightarrow C_{MCa\left(NO3\right)2}=\dfrac{0,1}{0,3}=0,3M\)

\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\\ 2Al+3CuCl_2\rightarrow2AlCl_3+3Cu\\ n_A=n_{AlCl_3}=n_{Al}=0,1\left(mol\right)\\ n_B=n_{Cu}=\dfrac{3}{2}.0,1=0,15\left(mol\right)\\ m_B=m_{Cu}=0,15.64=9,6\left(g\right)\\ C_{MddAlCl_3}=\dfrac{0,1}{0,2}=0,5\left(M\right)\)

PTHH: \(KOH+HCl\rightarrow KCl+H_2O\)

            \(KCl+AgNO_3\rightarrow KNO_3+AgCl\downarrow\)

Ta có: \(n_{HCl}=0,2\cdot2=0,4\left(mol\right)=n_{KOH}=n_{KNO_3}=n_{AgCl}\)

\(\Rightarrow\left\{{}\begin{matrix}m_{KOH}=0,1\cdot56=5,6\left(g\right)\\m_{AgCl}=0,1\cdot143,5=14,35\left(g\right)\\m_{KNO_3}=0,1\cdot101=10,1\left(g\right)\end{matrix}\right.\)

a) \(NaOH+HCl\rightarrow NaCl+H_2O\)

0,1................0,3

LẬp tỉ lệ : \(\dfrac{0,1}{1}< \dfrac{0,3}{1}\)=> Sau pứ HCl dư

\(m_{NaCl}=0,1.58,5=5,85\left(g\right)\)

b) \(CM_{NaCl}=\dfrac{0,1}{0,2+0,3}=0,2M\)

\(CM_{HCl\left(dư\right)}=\dfrac{\left(0,3-0,1\right)}{0,2+0,3}=0,4M\)

\(n_{AgNO3}=0,3\cdot1=0,3\) (mol)

n_HCl =0,5 . 0,5= 0,25 (mol)

_{AgNO3 + HCl -----> \(AgCl\downarrow+HNO3\)}

0,25<--- 0,25--->0,25---->0,25

=>\(n_{AgCl}=0,25\) (mol)

m_AgCl =0,25 . 143,5=35,875(g)

b)\(m_{dd_{ }sau_{ }pứ}\)=300 +500=800(ml)

m_HNO3 =0,25 . 63 =15,75(g)

C_%HNO3= \(\dfrac{15,75}{800}\cdot100\%=1,96\%\)

n_{AgNO3 dư} =0,3 - 0,25 =0,05 (mol)

m_{AgNO3 dư} =0,05 . 143,5=7,175(g)

\(C_{\%AgNO3_{ }dư}\)=\(\dfrac{7,175}{800}\cdot100\%=0,897\%\)

a)

$AgNO_3 + HCl \to AgCl + HNO_3$
Theo PTHH : 

$n_{AgCl} = n_{HCl} = n_{AgNO_3} = \dfrac{340.10\%}{170} =0,2(mol)$

$m_{dd\ HCl} = \dfrac{0,2.36,5}{7,3\%} = 100(gam)$

b)

$m_{AgCl} = 0,2.143,5 = 28,7(gam)$

c)

$m_{dd\ sau\ pư} = 340 + 100 -28,7 = 411,3(gam)$

$n_{HNO_3} = n_{AgNO_3} = 0,2(mol)$
$\Rightarrow C\%_{HNO_3} = \dfrac{0,2.63}{411,3}.100\% = 3,06\%$

cho cac axit :HCLO,HNO3,H2S,H2SO3,HNO2,HCLO4,HMno4.so axit manh la