Cho 10g dd HCl tác dụng với đ AgNO3 thì thu được 14.35g kết tủa. Tính C% của đ HCl phản ứng
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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)
$m_{dd\ sau\ pư} = 340 + 100 -28,7 = 411,3(gam)$
$n_{HNO_3} = n_{AgNO_3} = 0,2(mol)$
$\Rightarrow C\%_{HNO_3} = \dfrac{0,2.63}{411,3}.100\% = 3,06\%$
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Fe+2HCl\rightarrow FeCl_2+3H_2\\ Đặt:n_{Al}=a\left(mol\right);n_{Fe}=b\left(mol\right)\left(a,b>0\right)\\ \Rightarrow\left\{{}\begin{matrix}27a+56b=11\\1,5a+b=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\\ a,n_{HCl}=2.n_{H_2}=2.0,4=0,8\left(mol\right)\\ \Rightarrow V_{ddHCl}=\dfrac{0,8}{8}=0,1\left(l\right)\\ b,FeCl_2+2AgNO_3\rightarrow Fe\left(NO_3\right)_2+2AgCl\downarrow\\ AlCl_3+3AgNO_3\rightarrow Al\left(NO_3\right)_3+3AgCl\downarrow\\ n_{AgCl}=n_{AgNO_3}=3.n_{AlCl_3}+2.n_{FeCl_2}=3.a+2.b=3.0,2+2.0,1=0,8\left(mol\right)\\ \Rightarrow a=\dfrac{170.0,8}{250}.100=54,4\%\\ b=m_{\downarrow}=m_{AgCl}=0,8.143,5=114,8\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(MgCl_2+2NaOH\rightarrow2NaCl+Mg\left(OH\right)_{2\downarrow}\)
\(Mg\left(OH\right)_2\underrightarrow{t^o}MgO+H_2O\)
b, Ta có: \(n_{Mg}=\dfrac{9,6}{24}=0,4\left(mol\right)\)
Theo PT: \(n_{HCl}=2n_{Mg}=0,8\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}}=\dfrac{0,8}{0,2}=4\left(M\right)\)
c, Theo PT: \(n_{MgO}=n_{Mg}=0,4\left(mol\right)\)
\(\Rightarrow m_{MgO}=0,4.40=16\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a , \(nFe=\dfrac{11,2}{56}=0,2\left(mol\right)\)
, pthh:
\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
1mol 2mol 1mol 1mol
0,2 0,4 0,2 0,2
\(FeCl_2+2NaOH\rightarrow Fe\left(OH\right)_2\downarrow+2NaCl\)
1mol 2mol 1mol 2mol
0,2 0,4 0,2 0,4
b, \(mFe\left(OH\right)_2=0,2.90=18\left(gam\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
nglu=0,3 .suy ra:nag=0,6 mag=b=64,8
nagno3=nag+nagcl=0,6+0,15=0,75 magno3=127,5 a=(127,5/937,5).100=13,6
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Fe}=\dfrac{6,5}{56}=\dfrac{13}{112}mol\)
\(m_{CH_3COOH}=\dfrac{90\cdot20\%}{100\%}=18g\Rightarrow n_{CH_3COOH}=0,3mol\)
\(Fe+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Fe+H_2\uparrow\)
\(\dfrac{13}{112}\) 0,3 0 0
\(\dfrac{13}{112}\) \(\dfrac{13}{56}\) \(\dfrac{13}{112}\) \(\dfrac{13}{112}\)
0 \(\dfrac{19}{280}\) \(\dfrac{13}{112}\) \(\dfrac{13}{112}\)
a)\(m_{\left(CH_3COO\right)_2Fe}=\dfrac{13}{112}\cdot174=20,2g\)
\(m_{H_2}=\dfrac{13}{112}\cdot2=\dfrac{13}{56}g\)
\(m_{dd\left(CH_3COO\right)_2Fe}=6,5+90-\dfrac{13}{56}=96,27g\)
\(C\%=\dfrac{20,2}{96,27}\cdot100\%=20,98\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
=> \(C_{M_{HCl}}=\dfrac{0,25}{0,1}=2,5M\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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%
AgNO3 + HCl -> AgCl + HNO3 (1)
nAgCl=0,1(mol)
Từ 1:
nHCl=nAgNO3=0,1(mol)
mHCl=36,5.0,1=3,65(g)
C% dd HCl=\(\dfrac{3,65}{10}.100\%=36,5\%\)