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\(Đặt:n_{Na_2CO_3}=a\left(mol\right);n_{K_2CO_3}=b\left(mol\right)\\ Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O\\ K_2CO_3+2HCl\rightarrow2KCl+CO_2+H_2O\\ CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3\downarrow+H_2O\\ n_{CaCO_3}=n_{CO_2}=0,3\left(mol\right)\\ \Rightarrow\left\{{}\begin{matrix}106a+138b=38,2\\a+b=0,3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,2\end{matrix}\right.\\ a.C\%_{ddHCl}=\dfrac{0,6.36,5}{200}.100=10,95\%\\ b.m_{ddB}=38,2+200-0,3.44=225\left(g\right)\\ C\%_{ddKCl}=\dfrac{74,5.2.0,2}{225}.100\approx13,244\%\\ C\%_{ddNaCl}=\dfrac{58,5.2.0,1}{225}.100=5,2\%\)
a, \(MgCO_3+2HCl\rightarrow MgCl_2+CO_2+H_2O\)
Ta có: \(n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Theo PT: \(n_{MgCO_3}=n_{CO_2}=0,2\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{MgCO_3}=\dfrac{0,2.84}{64,8}.100\%\approx25,93\%\\\%m_{MgSO_4}\approx74,07\%\end{matrix}\right.\)
b, - Dung dịch C gồm: MgCl2, MgSO4 và HCl dư.
Theo PT: \(\left\{{}\begin{matrix}n_{MgCl_2}=n_{CO_2}=0,2\left(mol\right)\\n_{HCl\left(pư\right)}=2n_{CO_2}=0,4\left(mol\right)\end{matrix}\right.\)
Ta có: \(m_{HCl}=100.18,25\%=18,25\left(g\right)\Rightarrow n_{HCl}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\)
\(\Rightarrow n_{HCl\left(dư\right)}=0,5-0,4=0,1\left(mol\right)\)
\(m_{MgSO_4}=64,8-0,2.84=48\left(g\right)\Rightarrow n_{MgSO_4}=\dfrac{48}{120}=0,4\left(mol\right)\)
Có: m dd sau pư = 64,8 + 100 - 0,2.44 = 156 (g)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{MgCl_2}=\dfrac{0,2.95}{156}.100\%\approx12,18\%\\C\%_{HCl\left(dư\right)}=\dfrac{0,1.36,5}{156}.100\%\approx2,34\%\\C\%_{MgSO_4}=\dfrac{48}{156}.100\%\approx30,77\%\end{matrix}\right.\)
c, PT: \(MgCl_2+2NaOH\rightarrow2NaCl+Mg\left(OH\right)_{2\downarrow}\)
\(HCl+NaOH\rightarrow NaCl+H_2O\)
\(MgSO_4+2NaOH\rightarrow Na_2SO_4+Mg\left(OH\right)_{2\downarrow}\)
\(Mg\left(OH\right)_2\underrightarrow{t^o}MgO+H_2O\)
Theo PT: \(n_{MgO}=n_{Mg\left(OH\right)_2}=n_{MgCl_2}+n_{MgSO_4}=0,6\left(mol\right)\)
\(\Rightarrow m_{cr}=m_{MgO}=0,6.40=24\left(g\right)\)
PTHH: \(Na_2O+H_2O\rightarrow2NaOH\)
\(2NaOH+CuSO_4\rightarrow Na_2SO_4+Cu\left(OH\right)_2\downarrow\)
\(Cu\left(OH\right)_2\xrightarrow[]{t^o}CuO+H_2O\)
a) Ta có: \(n_{NaOH}=2n_{Na_2O}=2\cdot\dfrac{6,2}{62}=0,2\left(mol\right)\) \(\Rightarrow C\%_{NaOH}=\dfrac{0,2\cdot40}{6,2+193,8}\cdot100\%=4\%\)
b) Ta có: \(\left\{{}\begin{matrix}n_{NaOH}=0,2\left(mol\right)\\n_{CuSO_4}=\dfrac{200\cdot16\%}{160}=0,2\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,2}{2}< \dfrac{0,2}{1}\) \(\Rightarrow\) CuSO4 còn dư, tính theo NaOH
\(\Rightarrow n_{Cu\left(OH\right)_2}=0,1\left(mol\right)=n_{CuO}\) \(\Rightarrow m_{CuO}=0,1\cdot80=8\left(g\right)\)
c) PTHH: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)
Theo PTHH: \(n_{HCl}=2n_{CuO}=0,2\left(mol\right)\) \(\Rightarrow V_{ddHCl}=\dfrac{0,2}{2}=0,1\left(l\right)=100\left(ml\right)\)
a) Na2O +H2O-->2NaOH (1)
2NaOH +CuSO4 -->Na2SO4+ Cu(OH)2 (2)
Cu(OH)2 -to-> CuO +H2O (3)
b) mNa2O=8.100/100=8(g)
=>nNa2O=8/62=0,13(mol)
theo(2) :nCu(OH)2=1/2nNaOH=0,065(mol)
theo(3):nCuO=nCu(OH)2=0,065(mol)
=>mCuO=0,065.80=5,2(g)
c) CuO +2HCl-->CuCl2+H2O (4)
theo (4) : nHCl=2nCuO=0,13(mol)
mddHCl 25%=0,13.36,5.100250,13.36,5.10025=18,98(g)
MgO+H2SO4 → MgSO4+H2O(1)
MgCO3+H2SO4 → MgSO4+ CO2+H2O (2)
\(n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
Theo PT (2): nMgCO3=nCO2=0,1mol
=> mMgCO3=0,1.84=8,4g
mMgO=16,4-8,4=8g
=> nMgO=\(\dfrac{8}{40}\) = 0,2mol
Theo PT (1,2) ta có:nMgSO4=nMgO+nMgCO3=0,1+0,2=0,3 mol
nBa(OH)2=0,3.1,5 = 0,45 mol
Gọi x là số mol H2SO4 còn dư sau phản ứng (1,2).
H2SO4+Ba(OH)2→BaSO4↓+2H2O(3)
MgSO4+Ba(OH)2→BaSO4↓+Mg(OH)2↓(4)
Kết tủa tạo thành gồm BaSO4 và Mg(OH)2
Do đó dd B thu được là Ba(OH)2 dư
Ta có: 233x + 233.0,3 + 58.0,3 =110,6
=>x=0,1mol
Theo PT (3,4): nBa(OH)2 pứ=nH2SO4+nMgSO4=0,1+0,3=0,4mol
nBa(OH)2(dư)=0,45−0,4=0,05mol
CMBa(OH)2=\(\dfrac{0,05}{0,5}\)=0,1M
a. mdd X= 6,2+193,8= 200(g)=> C%X=\(\dfrac{6,2\cdot100}{200}=3,1\%\)
b. 2NaOH + CuSO4 ---> Na2SO4 + Cu(OH)2;
0,155--------0,0775------------0,0775----------0,0775 (mol)
Ta có: nCuSO4=\(\dfrac{200\cdot16}{100\cdot160}=0,2\left(mol\right)\)
nNaOH=\(\dfrac{6,2}{40}0,155\left(mol\right)\)
Xét tỉ lệ:\(\dfrac{nNaOH}{nNaOHpt}=\dfrac{0,155}{2}< \dfrac{nCuSO4}{nCuSO4pt}=\dfrac{0,2}{1}\)
=> CuSO4 dư. Sản phẩm tính theo NaOH.
=> nNa2SO4=0,155/2= 0,0775(mol)=> mNa2SO4=0,0775*142=11,005(g).
nCu(OH)2=0,155/2=0,0775(mol)=> mCu(OH)2=0,0775*98=7,595(g).
=> mdd sau pư= 200+200-7,595=392,405(g)
=> C%ddA=\(\dfrac{11,005\cdot100}{392,405}=2,8\%\)
c. Cu(OH)2 ---to-> CuO + H2O;
ta có: nCu(OH)2=0,0775(mol)=> nCuO=0,0775(mol)
CuO + 2 HCl --> CuCl2 + H2O;
0,0775---0,155 (mol)
nHCl=0,155(mol)=>mHCl=0,155*36,5=5,6575(g)
Ba+2H2O->Ba(OH)2+H2
0,074-----------0,074 mol
Ba(OH)2+CuSO4->Cu(OH)2+BaSO4
n Ba=\(\dfrac{17}{137}=0,124mol\)
m CuSO4=8g
n CuSO4=\(\dfrac{8}{160}=0,05mol\)
->Ba dư
=>n Ba(OH)2dư=0,124-0,05=0,074 mol
C% =\(\dfrac{0,074.171}{17+200}.100=5,8\%\)