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![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{MgCl_2}\)=\(0,1.2=0,2(mol)\)
\(n_{Ba(OH)_2}\)=\(0,15.1,5=0,225(mol) \)
\({MgCl_2}+{Ba(OH)_2}-->{Mg(OH)_2}+{BaCl_2}\)
Dung dịch A chứa 0,225-0,2=0,025 mol \({Ba(OH)_2}\) dư; 0,2 mol \({BaCl_2}\)
Kết tủa B là 0,2 mol \({Mg(OH)_2}\)
\({Mg(OH)_2}-->MgO+{H_2O}\)
⇒\(n_{MgO}\)=\(n_{Mg(OH)_2}=0,2 mol\)
⇒\(m_{MgO}=0,2.40=8(g)\)
Coi thể tích dung dịch không đổi sau khi trộn
\(V_{dd}=100+150=250ml=0,25l\)
⇒\(C_M{Ba(OH)_2}\)=\(\dfrac{0,025}{0,25}=0,1M\)
\(C_M{BaCl_2}=\dfrac{0,2}{0,25}=0,8M\)
mdd(sau phản ứng)=250.1,12=280(g)
C%\({Ba(OH)_2}=\dfrac{0,025.171}{280}.100=1,5%\)%
C%\({BaCl_2}=\dfrac{0,2.208}{280}.100=14,85%\)%
![](https://rs.olm.vn/images/avt/0.png?1311)
mdd NaOH = 62,5.1,12 = 70 (g)
=> \(n_{NaOH}=\dfrac{70.16\%}{40}=0,28\left(mol\right)\)
Gọi \(\left\{{}\begin{matrix}C_{M\left(H_2SO_4\right)}=aM\\C_{M\left(Cu\left(NO_3\right)_2\right)}=bM\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}n_{H_2SO_4}=0,1a\left(mol\right)\\n_{Cu\left(NO_3\right)_2}=0,1b\left(mol\right)\end{matrix}\right.\)
PTHH: 2NaOH + H2SO4 --> Na2SO4 + 2H2O
0,2a<----0,1a
2NaOH + Cu(NO3)2 --> Cu(OH)2 + 2NaNO3
0,2b<-----0,1b--------->0,1b
Cu(OH)2 --to--> CuO + H2O
0,1b------------>0,1b
=> \(0,1b=\dfrac{1,6}{80}=0,02\)
=> b = 0,2
Có: nNaOH = 0,2a + 0,2b = 0,28
=> a = 1,2
Vậy \(\left\{{}\begin{matrix}C_{M\left(H_2SO_4\right)}=1,2M\\C_{M\left(Cu\left(NO_3\right)_2\right)}=0,2M\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)PTHH: \(Ba\left(OH\right)+Na_2CO_3\rightarrow2NaOH+BaCO_3\downarrow\)
\(BaCO_3\underrightarrow{t^o}BaO+CO_2\uparrow\)
Ta có: \(n_{Ba\left(OH\right)_2}=0,4\cdot0,2=0,08\left(mol\right)\)
\(\Rightarrow n_{NaOH}=0,16mol\) \(\Rightarrow C_{M_{NaOH}}=\dfrac{0,16}{0,4}=0,4\left(M\right)\) (Coi Vdd thay đổi không đáng kể)
b) Theo PTHH: \(n_{BaCO_3}=n_{Ba\left(OH\right)_2}=n_{BaO}=0,08mol\) \(\Rightarrow m_{BaO}=0,08\cdot153=12,24\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a. Ba(OH)2 +Na2CO3 ➝ BaCO3 + 2NaOH
BaCO3 ➝ BaO + CO2
nBa(OH)2 = 0,08 mol
=> nNaOH = 2nBa(OH)2 = 0,16 mol
=> CM = 0,4 M
b) Bảo toàn Ba: nBaO = nBa(OH)2 = 0,08 mol
=> m = 12,24 g
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
$FeCl_3 + 3NaOH \to Fe(OH)_3 + 3NaCl$
$2Fe(OH)_3 \xrightarrow{t^o} Fe_2O_3 + 3H_2O$
$n_{FeCl_3} = 0,3.2 = 0,6(mol)$
Theo PTHH : $n_{Fe_2O_3} = \dfrac{1}{2}n_{FeCl_3} = 0,3(mol)$
$\Rightarrow m_{Fe_2O_3} = 0,3.160 = 48(gam)$
b) Sau phản ứng, $V_{dd} = 0,3 + 0,3 = 0,6(lít)$
$n_{NaCl} = 3n_{FeCl_3} = 1,8(mol) \Rightarrow C_{M_{NaCl}} = \dfrac{1,8}{0,6} = 3M$
![](https://rs.olm.vn/images/avt/0.png?1311)
a)2NaOH+H2SO4→Na2SO4+2H2O(1)
Cu(NO3)2+2NaOH→Cu(OH)2+2NaNO3(2)
Cu(OH)2→CuO+H2O(3)
nCuO=\(\dfrac{1,6}{80}\)=0,02mol
mddNaOH=31,25×1,12=35g
nNaOH=35×16%40=0,14mol
nNaOH(2)=0,02×2=0,04mol
⇒nNaOH(1)=0,14−0,04=0,1mol
nH2SO4=0,12=0,05mol
CM(H2SO4)=\(\dfrac{0,05}{0,05}\)=1M
CM(Cu(NO3)2)=\(\dfrac{0,02}{0,05}\)=0,4M
b)nCu=\(\dfrac{2,4}{64}\)=0,0375mol
nH+=2nH2SO4=0,1mol
nNO3−=2nCu(NO3)2=0,04mol
Cu+4H++NO3−→Cu2++NO+2H2O
\(\dfrac{0,04}{1}\)>\(\dfrac{0,03751}{1}\)>\(\dfrac{0,1}{4}\)⇒ Tính theo ion H+nNO=0,14=0,025mol
⇒VNO=0,025×22,4=0,56l
![](https://rs.olm.vn/images/avt/0.png?1311)
PTHH: \(CuCl_2+2NaOH\rightarrow2NaCl+Cu\left(OH\right)_2\downarrow\)
\(Cu\left(OH\right)_2\xrightarrow[]{t^o}CuO+H_2O\)
Ta có: \(\left\{{}\begin{matrix}n_{CuCl_2}=0,2\cdot2=0,4\left(mol\right)\\n_{NaOH}=0,2\cdot2=0,4\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,4}{1}>\dfrac{0,4}{2}\) \(\Rightarrow\) CuCl2 còn dư
\(\Rightarrow\left\{{}\begin{matrix}n_{NaCl}=0,4\left(mol\right)\\n_{Cu\left(OH\right)_2}=0,2\left(mol\right)=n_{CuO}=n_{CuCl_2\left(dư\right)}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{CuO}=0,2\cdot80=16\left(g\right)\\C_{M_{NaCl}}=\dfrac{0,4}{0,2+0,2}=1\left(M\right)\\C_{M_{CuCl_2}}=\dfrac{0,2}{0,4}=0,5\left(M\right)\\\end{matrix}\right.\)
1.
2.
a)
+nFe2(SO4)3 = 0.1*2 = 0.2 (mol)
+nBa(OH)2 = 0.15*1.5 = 0.225 (mol)
3Ba(OH)2 + Fe2(SO4)3 => 2Fe(OH)3↓ + 3BaSO4↓(1)
0.225...................0.2.................
2Fe(OH)3(t*) => Fe2O3 + 3H2O(2)
0.15.........................0.075...........
_Dựa vào phương trình (1) ta thấy Fe2(SO4)3 còn dư 0.125 mol => dd(B) : Fe2(SO4)3
Fe2(SO4)3 + 3BaCl2 => 3BaSO4↓ + 2FeCl3
0.125..................0.375............0.375
b)
_Chất rắn (D) : Fe2O3 và BaSO4 không bị phân hủy.
=>m(D) = mFe2O3 + mBaSO4 = 0.075*160 + 0.375*233 = 99.375(g)
_Chất rắn (E) : BaSO4
=>m(E) = mBaSO4 = 0.375*233 = 87.375(g)
c)
_Dung dịch (B) : Fe2(SO4)3
=>Vdd(sau) = 150 + 100 = 250 (ml) = 0.25 (lit)
=>nFe2(SO4)3 (dư) = 0.125 (mol)
=>CM(Fe2(SO4)3) = 0.125 / 0.25 = 0.5 (M)