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

0,1                   0,1               0,1               0,1            mol

\(n_{NaCl}=\frac{\frac{5,85}{100}.100}{58,5}=0,1mol\)

\(n_{AgNO_3}=\frac{\frac{17}{100}.200}{170}=0,2mol\)

Vậy NaCl hết và \(AgNO_3\) dư

a. \(m_{AgCl}=0,1.143,5=14,35g\)

b. Chất tan của dd X là: \(NaNO_3;AgNO_{3\left(dư\right)}\)

\(m_{ddNaNO_3}=100+200-14,35=285,65g\)

\(C\%_{NaNO_3}=\frac{0,1.85}{285,65}.100\approx2,98\%\)

\(C\%_{AgNO_3\left(dư\right)}=\frac{\left(0,2-0,1\right).170}{285,65}.100\approx5,95\%\)

$NaCl+AgN{O}_3\to AgCl\downarrow +NaN{O}_3$

0,1                   0,1               0,1               0,1            mol

$n_{NaCl}=\frac{\frac{5,85}{100}.100}{58,5}=0,1mol$

$n_{AgN{O}_3}=\frac{\frac{17}{100}.200}{170}=0,2mol$

Vậy NaCl hết và $AgN{O}_3$ dư

a. $m_{AgCl}=0,1.143,5=14,35g$

b. Chất tan của dd X là: $NaN{O}_3;AgN{O}_{3\left(dư\right)}$

$m_{ddNaN{O}_3}=100+200-14,35=285,65g$

$C{\%}_{NaN{O}_3}=\frac{0,1.85}{285,65}.100\approx 2,98\%$

$C{\%}_{AgN{O}_3\left(dư\right)}=\frac{\left(0,2-0,1\right).170}{285,65}.100\approx 5,95\%$

$\mathrm{Mik\; giải\; giống\; bạn\; Nam\; nhé}$

$\mathrm{HT}$

\(a,C\%_{NaCl}=\dfrac{15}{15+185}.100\%=7,5\%\\ b,m_{HNO_3}=\dfrac{18,9}{100}.100+\dfrac{6,3}{100}.200=31,5\left(g\right)\\ m_{ddHNO_3}=100+200=300\left(g\right)\\ C\%_{HNO_3}=\dfrac{31,5}{300}.100\%=10,5\%\)

\(c,n_{NaCl}=\dfrac{5,85}{58,5}=0,1\left(mol\right)\\ C_{M\left(NaCl\right)}=\dfrac{0,1}{0,1}=1M\\ d,n_{KOH}=2.0,2+0,2.0,2=0,44\left(mol\right)\\ V_{ddKOH}=0,2+0,2=0,4\left(l\right)\\ C_{M\left(KOH\right)}=\dfrac{0,44}{0,4}=1,1M\\ e,m_{NaOH}=\dfrac{150.16}{100}=24\left(g\right)\\ m_{ddNaOH}=50+150=200\left(g\right)\\ C\%_{NaOH}=\dfrac{24}{200}.100\%=12\%\)

a)

NaCl + AgNO3 -> NaNO3 + AgCl↓

0,1 0,1 0,1 0,1

b)

mNaCl = \(\frac{100.5,85}{100}=5,85\)=> nNaCl = 5,85 : 58,5 = 0,1 mol

mAgCl = 0,1 . 143,5 = 14,35 g

c)

mNaNO3 = 0,1 . 85 = 8,5 g

a, NaCl + AgNO₃ --> NaNO₃ +AgCl

1mol 1mol 1mol 1mol

0,1mol 0,1mol 0,1mol 0,1mol

b, m_NaCl=100.5,85%=5,85g

n_NaCl=\(\frac{5,85}{58,5}\)=0,1(mol)

m_AgNO3=0,1.170=17g

m_NaNO3=0,1.85=8,5g

m_AgCl=0,1.143,5=14,35g

a. PTHH : NaCl + AgNO3 ---> NaNO3 + AgCl.
b. mNaCl = ( 5,85 . 100 ) : 100 = 5,85 ( g )
nNaCl = 5,85 : 58,5 = 0,1 ( mol )
PTHH : NaCl + AgNO3 ---> NaNO3 + AgCl.
Theo PT : 1mol 1 mol 1 mol 1 mol
Theo Bài: 0,1 mol 0,1 mol 0,1 mol 0,1 mol
mAgCl = 0,1 . 143,5 = 14,35 ( g )
c. mNaNO3 = 0,1 . 85 = 8,5 ( g )
mAgNO3 = 0,1 . 170 = 17 ( g )

 

\(m_{NaCl}=\dfrac{5,85\%.200}{100\%}=11,7\left(g\right)\)
\(n_{NaCl}=\dfrac{11,7}{58,5}=0,2\left(mol\right)\)

nNaCl=0,2 mol nha bạn

ta có C%=mct/mdd.100

=>mct=C%.mđ/100

thay:  5,85.200/100=11,7g

nNaCl=m/M thay 11,7/58,5 =0.2 mol

\(m_{NaCl}=\dfrac{5,85.100}{100}=5,85g\)

\(n_{NaCl}=\dfrac{5,85}{58,5}=0,1\left(mol\right)\)

\(m_{AgNO_3}=\dfrac{12.100}{100}=12g\)

\(n_{AgNO_3}=\dfrac{12}{170}\approx0,07\left(mol\right)\)

Ta co: \(\dfrac{0,1}{1}>\dfrac{0,07}{1}\Rightarrow\) NaCl dư

NaCl + AgNO₃ \(\rightarrow\) NaNO₃ + AgCl\(\left(\downarrow\right)\)

de: 0,1 0,07

pu: 0,07 0,07 0,07 0,07

spu: 0,03 0 0,07 0,07

\(m_{NaCl\left(dư\right)}=0,03.58,5=1,755g\)

\(m_{NaNO_3}=0,07.85=5,95g\)

\(m_{ddspu}=m_{ddNaCl}+m_{ddAgNO_3}-m_{AgCl}=189,955g\)

\(C\%_{NaCl\left(dư\right)}=\dfrac{1,755}{189,955}.100\%\approx0,92\%\)

\(C\%_{NaNO_3}=\dfrac{5,95}{189,955}.100\%\approx3,13\%\)

Câu 1:

a) \(2KMnO_4+16HCl\rightarrow2KCl+2MnCl_2+5Cl_2+8H_2O\)

    \(\dfrac{1}{2}Cl_2+K\underrightarrow{t^o}KCl\)

     \(2KCl+2H_2O\xrightarrow[cómàngngăn]{đp}2KOH+Cl_2+H_2\)

     \(Cl_2+H_2O⇌HCl+HClO\)

     \(NaOH+HClO\rightarrow NaClO+H_2O\)

     \(NaClO_{\left(rắn\right)}+HCl\rightarrow NaCl+Cl_2+H_2O\)

     \(2NaCl+2H_2O\xrightarrow[cómàngngăn]{đp}2NaOH+Cl_2+H_2O\)

     \(3Cl_2+2Fe\underrightarrow{t^o}2FeCl_3\)

b) \(MnO_2+4HCl\rightarrow MnCl_2+Cl_2+2H_2O\)

    \(Cl_2+H_2\underrightarrow{a/s}2HCl\)

    \(2HCl+Fe\rightarrow FeCl_2+H_2\)

    \(2FeCl_2+Cl_2\rightarrow2FeCl_3\)

    \(FeCl_3+3KOH\rightarrow3KCl+Fe\left(OH\right)_3\downarrow\)

    \(2KCl+2H_2O\xrightarrow[cmn]{đp}2KOH+Cl_2+H_2\)

    \(6KOH+3Cl_2\underrightarrow{t^o}5KCl+KClO_3+3H_2O\)

    \(KClO_3+6HCl\rightarrow KCl+3Cl_2+3H_2O\)

1)\(n_{NaOH}:\dfrac{60.10\%}{100\%.40}=0,15\left(mol\right)\)

KL dung dịch sau p/ư: 60+40=100(g)

\(n_{NaCl}:\dfrac{100.5,85\%}{100\%.58,5}=0,1\left(mol\right)\)

\(NaOH+HCl\rightarrow NaCl+H_2O\)

1...................1...............1.................(mol)

0,1................0,1............0,1...............(mol)

-> NaOH dư

C% dd HCl: \(\dfrac{0,1.36,5}{40}.100\%=9,125\%\)

Đáp án C.