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a,
PTHH: CuO + 2HCl → CuCl2 + H2
Mol: x x
PTHH: NaOH + HCl → NaCl + H2O
Mol: y y
Ta có:\(\left\{{}\begin{matrix}80x+40y=10\\135x+58,5y=16,5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,104\\y=0,042\end{matrix}\right.\)
PTHH: CuO + 2HCl → CuCl2 + H2
Mol: 0,104 0,208
PTHH: NaOH + HCl → NaCl + H2O
Mol: 0,042 0,042
\(\Rightarrow\%m_{CuO}=\dfrac{0,104.80.100}{10}=83,2\%;\%m_{NaOH}=100\%-83,2\%=16,8\%\)
b,\(n_{HCl}=0,208+0,042=0,25\left(mol\right)\)
\(\Rightarrow C_{M_{ddHCl}}=\dfrac{0,25}{0,2}=1,25M\)
\(Fe+2HCl\rightarrow FeCl2+H2\)(1)
1____2________1_______1
\(CuO+2HCl\rightarrow CuCl2+H2O\)(2)
1_______2__________1_____1
Ta có :\(n_{H2}=\frac{3,36}{22,4}=0,15\left(mol\right)\)
\(n_{Fe}=\frac{0,15.1}{1}=0,15\left(mol\right)\)
\(\rightarrow m_{Fe}=0,15.56=8,4g\)
\(\rightarrow m_{CuO}=10-8,4=1,6g\)
b,Theo PT(1)\(n_{HCl}=\frac{0,15.2}{1}=0,3\left(mol\right)\)
\(\rightarrow n_{CuO}=\frac{1,6}{80}=0,02\left(mol\right)\)
Theo PT(2)\(n_{HCl}=\frac{0,02.2}{1}=0,04\left(mol\right)\)
\(\rightarrow\Sigma n_{HCl}=0,3+0,04=0,34\left(mol\right)\)
\(\rightarrow CM_{HCl}=\frac{0,34}{0,2}=1,7M\)
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: \(n_{MgO}=n_{Mg}=0,4\left(mol\right)\)
\(\Rightarrow m_{MgO}=0,4.40=16\left(g\right)\)
PTHH
Mg + 2HCl ----> MgCl2 + H2 (1)
MgO + 2HCl -----> MgCl2 + H2O (2)
a) Theo pt(1) n Mg = n H2 = \(\frac{1,12}{22,4}\) = 0,05 (mol)
==> m Mg = 0,005 . 24=1,2 (g)
%m Mg = \(\frac{1,2}{3,2}\). 100%= 37,5%
%m MgO= 100% - 37,5%= 62,5%
b)m dd sau pư = 3,2 + 246,9 - 0,05 . 2=250 (g)
Theo pt(1)(2) n MgCl2(1) = n Mg = 0,05 mol
n MgCl2 (2) = n MgO=\(\frac{3,2-1,2}{40}\)=0,05(mol)
==> tổng n MgCl2 = 0,1 (mol) ---->m MgCl2 = 9,5 (g)
C%(MgCl2)= \(\frac{9,5}{250}\) .100% = 3,8%
1. Fe+2HCl-> FeCl2 + H2 (1)
CuO + 2HCl -> CuCl2 + H2O(2)
a.nH2=0,02(mol)=>nFe=0,02(mol)=> mFe=1,12(g)
=> nHCl(1)=0,04(mol)
nHCl dùng=\(\dfrac{15\cdot14,6}{100}=2,19\left(g\right)\)
=> nHCl=0,06(mol)
=>nHCl(2)=0,02(mol)=> nCuO=0,01(mol)
=> mCuO=0,8(g)
b. mddA=1,12+0,8+15-0,02*2=16,88(g)
mFeCl2=0,02*127=2,54(g)
mCuCl2=0,01*135=1,35(g)
C% FeCl2=\(\dfrac{2,54\cdot100}{16,88}=15,047\%\)
C% CuCl2=\(\dfrac{1,35\cdot100}{16,88}=7,9976\%\)
Bài 2
mol HCl=3.0,1=0,3mol(100ml=0,1l)
CuO+2HCl->CuCl2+H2O (1)
xmol 2xmol
ZnO+2HCl->ZnCl2+H2O(2)
ymol 2ymol
Từ 1 và 2 ta co hệ phương trình
2x+2y=0,3 ->x=0,05=molCuO
80x+81y=12,1 ->y=0,1=molZnO
=>mCuO=0,05.80=4g
->%CuO=(4.100)/12,1=33,075%
->%ZnO=100-33,075=66,943%
b. CuO+H2SO4->CuSO4+H2O (3)
Theo ptpu 3 taco nH2SO4=nCuO=0,05 mol
ZnO+H2SO4->ZnSO4+H2O (4)
Theo ptpu 4 ta co nH2SO4=nZnO=0,1mol
=>nH2SO4=0.05+0,1=0,15mol
->mH2SO4=0,15.98=14,7g
=>mddH2SO4=(14,7.100)/20=73,5g
Bài 1
a/. Phương trình phản ứng hoá học:
Fe + 2HCl --> FeCl2 + H2
b/. nH2 = V/22,4 = 3,36/22,4 = 0,15 (mol)
....... Fe.....+ 2HCl --> Fecl2 + H2
TPT 1 mol....2 mol.................1 mol
TDB x mol....y mol................0,15 mol
nFe = x = (0,15x1)/1 = 0,15 (mol)
mFe = n x M = 0,15 x 56 = 8,4 (g)
c/. nHCl = y = (0,15x2)/1 = 0,3 (mol)
CMHCl = n/V = 0,3/0,05 = 6 (M)