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)

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

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

PTHH: \(CaCl_2+Na_2CO_3\rightarrow2NaCl+CaCO_3\downarrow\)

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

A là dd NaCl

B là CaCO₃

C là dd NaNO₃

D là AgCl

Ta có: \(n_{AgCl}=\dfrac{57,4}{143,5}=0,4\left(mol\right)=n_{NaCl}\)

\(\Rightarrow n_{CaCO_3}=0,2\left(mol\right)=n_{Na_2CO_3}\) \(\Rightarrow\left\{{}\begin{matrix}m_{NaCl}=0,4\cdot58,5=23,4\left(g\right)\\m_{CaCO_3}=0,2\cdot100=20\left(g\right)\\m_{ddNa_2CO_3}=\dfrac{0,2\cdot106}{16\%}=132,5\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{ddA}=m_{ddCaCl_2}+m_{ddNa_2CO_3}-m_{CaCO_3}=290,1\left(g\right)\)

\(\Rightarrow C\%_{NaCl}=\dfrac{23,4}{290,1}\cdot100\%\approx8,07\%\)

Bài 1 ⅕ dd A có mH2O=92,7/5=18,54g

HCl         + AgNO3 =>AgCl       + HNO3

0,04 mol<=0,04 mol<=0,04 mol

nAgCl=5,74/143,5=0,04 mol

mHCl=0,04.36,5=1,46g

mddHCl=0,04.36,5+18,54=20g

C%dd HCl=C%dd A=1,46/20.100%=7,3%

Bài 2

TH1 2 Halogen là F và Cl

n ktủa=nAgCl=4,75/143,5=0,033 mol>0,03 mol hh 2 muối

=>loại

TH2 2 Halogen không phải F và Cl

Gọi Hal trung bình là Z

nAgZ=4,75/(108+Z) mol

nhh muối=0,03 mol

NaZ + AgNO3 =>AgZ + NaNO3

0,03 mol       =>0,03 mol

=>4,75/0,03=108+Z=>Z=50,33g/mol

Mà 2 halogen ở 2 chu kì kế tiếp nhau

=>2 halogen là Cl và Br

nAgNO3= (100.17%)/170=0,1(mol)

nHCl= (300.3,65%)/36,5=0,3(mol)

a) PTHH: AgNO3 + HCl -> AgCl + HNO3

Ta có: 0,1/1 < 0,3/1

=> AgNO3 hết, HCl dư, tính theo nAgNO3

Ta có: nAgCl= nHNO3= nHCl(p.ứ)= nAgNO3= 0,1(mol)

=>m(kt)=mAgCl= 143,5.0,1= 14,35(g)

b) mHCl(dư)= (0,3- 0,1).36,5=7,3(g)

mHNO3= 63.0,1= 6,3(g)

mddsau= mddAgNO3 + mddHCl - mAgCl= 100+300- 14,35= 385,65(g)

=>C%ddHCl(dư)= (7,3/385,65).100= 1,893%

C%ddHNO3= (6,3/385,65).100=1,634%

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

Đáp án C

+) ½ X +  BaCl₂:

Ba 2 +   +   CO 3 2 -   →   BaCO 3 ↓                               0 , 06                         0 , 06

+) ½ X +  CaCl₂:

2 HCO 3 -       →   CO 2 ↑   +   H 2 O   + CO 3 2 -       x 2                                                                                                   0 , 25 x Ca 2 +     +         CO 3 2 -   →         CaCO 3 ↓ ( 0 , 25 x   +   0 , 06 )             → ( 0 , 25 x   + 0 , 06 ) = 0 , 07 HCO 3 -   +   OH -   → CO 3 2 -   + H 2 O 0 , 12                 ← 0 , 12     ← 0 , 12

⇒ tổng  nHCO₃^- =  0,12 + 0,04 = 0,06 mol

⇒ a = 0,08 mol/l

⇒ m = 40.0,12 = 4,8 g

HCl + AgNO₃ ➜ AgCl↓ + HNO₃

\(n_{HCl}=0,2\times2=0,4\left(mol\right)\)

\(n_{AgNO_3}=0,3\times2=0,6\left(mol\right)\)

Theo PT: \(n_{HCl}=n_{AgNO_3}\)

Theo bài: \(n_{HCl}=\dfrac{2}{3}n_{AgNO_3}\)

Vì \(\dfrac{2}{3}< 1\) ⇒ dd HCl hết, dd AgNO₃ dư

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

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

Dung dịch B gồm: AgNO₃ dư và HNO₃

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

\(\Rightarrow n_{AgNO_3}dư=0,6-0,4=0,2\left(mol\right)\)

Theo PT: \(n_{HNO_3}=n_{HCl}=0,4\left(mol\right)\)

\(\Sigma n_{ctB}=n_{AgNO_3}dư+n_{HNO_3}=0,2+0,4=0,6\left(mol\right)\)

\(\Sigma m_{ddB}=0,2+0,3=0,5\left(l\right)\)

\(\Rightarrow C_{M_{ddB}}=\dfrac{0,6}{0,5}=1,2\left(M\right)\)

HCl + AgNO₃ ➜ AgCl↓ + HNO₃

\(n_{HCl}=0,2\times2=0,4\left(mol\right)\)

\(n_{AgNO_3}=0,3\times2=0,6\left(mol\right)\)

Theo PT: \(n_{HCl}=n_{AgNO_3}\)

Theo bài: \(n_{HCl}=\dfrac{2}{3}n_{AgNO_3}\)

Vì \(\dfrac{2}{3}< 1\) ⇒ dd HCl hết, dd AgNO₃ dư

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

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

Dung dịch B gồm: AgNO₃ dư và HNO₃

\(\Sigma m_{ddB}=0,2+0,3=0,5\left(l\right)\)

Theo PT: \(n_{HNO_3}=n_{HCl}=0,4\left(mol\right)\)

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

\(\Rightarrow n_{AgNO_3}dư=0,6-0,4=0,2\left(mol\right)\)

\(\Sigma n_{ctB}=n_{AgNO_3}dư+n_{HNO_3}=0,2+0,4=0,6\left(mol\right)\)

\(\Rightarrow C_{M_{ddB}}=\dfrac{0,6}{0,5}=1,2\left(M\right)\)

a, \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

\(FeCl_3+3KOH\rightarrow3KCl+Fe\left(OH\right)_{3\downarrow}\)

\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)

Theo PT: \(n_{HCl}=6n_{Fe_2O_3}=0,6\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,6.36,5=21,9\left(g\right)\)

b, \(n_{Fe\left(OH\right)_3}=n_{FeCl_3}=2n_{Fe_2O_3}=0,2\left(mol\right)\)

\(\Rightarrow m_{Fe\left(OH\right)_3}=0,2.107=21,4\left(g\right)\)

\(n_{KOH}=3n_{FeCl_3}=0,6\left(mol\right)\)

\(\Rightarrow C_{M_{KOH}}=\dfrac{0,6}{0,2}=3\left(M\right)\)