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

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

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

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)

$AgNO_3 + HCl \to AgCl + HNO_3$
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: \(NaCl+AgNO_3\rightarrow NaNO_3+AgCl\downarrow\)

Ta có: \(n_{NaCl}=0,2\cdot0,5=0,1\left(mol\right)=n_{AgNO_3}=n_{AgCl}\)

\(\Rightarrow\left\{{}\begin{matrix}C_{M_{AgNO_3}}=\dfrac{0,1}{0,4}=0,25\left(M\right)\\m_{AgCl}=0,1\cdot143,5=14,35\left(g\right)\end{matrix}\right.\)

a) nNaCl=0,4.2=0,8(mol)

nAgNO3= 2.0,6=1,2(mol)

PTHH: NaCl + AgNO3 -> NaNO3 + AgCl (kết tủa)

Ta có: nNaCl(p.ứ)/nNaCl(PT) = 0,8/1 < nAgNO3(p.ứ)/nAgNO3(pt)= 1,2/1

=> P.ứ dư AgNO3, có tạo kt AgCl.

nAgCl= nNaNO3=nAgNO3(P.ứ)=nNaCl= 0,8(mol)

=> m(kết tủa)= mAgCl=0,8. 143,5= 114,8(g)

b) Vddsau= 400+600=1000(ml)=1(l)

CMddNaNO3=0,8/1=0,8(M)

CMddAgNO3(dư)= (1,2-0,8.1)/1= 0,4(M)

số xấu lắm bn coi có lộn số j hơm:<

nAl=0,337 mol á 

t có đề i hệt giải cho dc khong

\(a,PTHH:2Al+3X_2\rightarrow2AlX_3\\ Theo.ĐLBTKL,ta.có:\\ m_{Al}+m_{X_2}=m_{AlX_3}\\ \Leftrightarrow m_{Al}+33,6=37,38\\ \Leftrightarrow m_{Al}=3,78\left(g\right)\\ \Rightarrow n_{Al}=\dfrac{3,78}{27}=0,14\left(mol\right)\\ n_{X_2}=\dfrac{3}{2}.0,14=0,21\left(mol\right)\\ \Rightarrow M_{X_2}=\dfrac{33,6}{0,21}=160\left(\dfrac{g}{mol}\right)\\ \Rightarrow M_X=80\left(\dfrac{g}{mol}\right)\\ \Rightarrow X:Brom\left(Br=80\right)\\ b,n_{AlBr_3}=\dfrac{42,72}{267}=0,16\left(mol\right)\\ AlBr_3+3AgNO_3\rightarrow Al\left(NO_3\right)_3+3AgBr\downarrow\left(vàng.nhạt\right)\\ n_{AgBr}=n_{AgNO_3}=0,16.3=0,48\left(mol\right)\\ \Rightarrow m_T=m_{\downarrow}=m_{AgBr}=188.0,48=90,24\left(g\right)\)

\(m_{ddZ}=42,72+447,52-90,24=400\left(g\right)\\ n_{Al\left(NO_3\right)_3}=n_{AlBr_3}=0,16\left(mol\right)\\ \Rightarrow C\%_{ddZ}=C\%_{ddAl\left(NO_3\right)_3}=\dfrac{0,16.213}{400}.100=8,52\%\)

\(c,n_{Br_2}=\dfrac{12,8}{160}=0,08\left(mol\right)\\ 2NaI+Br_2\rightarrow2NaBr+I_2\\ n_{I_2}=n_{Br_2}=0,08\left(mol\right);n_{NaI}=2.0,08=0,16\left(mol\right)\\ \Rightarrow x=C_{MddNaI}=\dfrac{0,16}{0,25}=0,64\left(M\right)\\ m_{I_2}=0,08.254=20,32\left(g\right)\)