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a) \(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)
b) \(n_{CuSO_4}=0,4.1=0,4\left(mol\right)\)
\(n_{NaOH}=\dfrac{40}{40}=1\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,4}{1}< \dfrac{1}{2}\) => CuSO4 hết, NaOH dư
PTHH: CuSO4 + 2NaOH --> Cu(OH)2 + Na2SO4
_______0,4----->0,8---------->0,4-------->0,4
Cu(OH)2 --to--> CuO + H2O
0,4------------->0,4
=> mCuO = 0,4.80 = 32 (g)
c) \(\left\{{}\begin{matrix}m_{NaOH\left(dư\right)}=\left(1-0,8\right).40=8\left(g\right)\\m_{Na_2SO_4}=0,4.142=56,8\left(g\right)\end{matrix}\right.\)
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{HCl}=0,2.2=0,4\left(mol\right)\\ m_{ddHCl}=\dfrac{0,4.36,5.100}{10}=146\left(g\right)\\ n_{H_2}=n_{ZnCl_2}=n_{Zn}=0,2\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\\ C\%_{ddZnCl_2}=\dfrac{136.0,2}{13+146-0,2.2}.100\approx17,15\%\)
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
PTHH: Zn + 2HCl → ZnCl2 + H2
Mol: 0,2 0,4 0,2 0,2
\(V_{H_2}=0,2.22,4=4,48\left(l\right)\)
\(m_{ddHCl}=\dfrac{0,4.36,5.100}{10}=146\left(g\right)\)
mdd sau pứ = 13+146-0,2.2 = 158,6 (g)
\(C\%_{ddZnCl_2}=\dfrac{0,2.136.100\%}{158,6}=17,15\%\)
a)
$Zn + CuSO_4 \to ZnSO_4 + Cu$
b)
Theo PTHH : $n_{Zn} = n_{CuSO_4} = \dfrac{3,2.10\%}{160} = 0,002(mol)$
$m_{Zn} = 0,002.65 = 0,13(gam)$
c)
$n_{Cu} = 0,002(mol)$
$\Rightarrow m_{dd\ sau\ pư} = 0,13 + 3,2 - 0,002.64 = 3,202(gam)$
$C\%_{ZnSO_4} = \dfrac{0,002.161}{3,202}.100\% = 10,06\%$
a)
$Mg + H_2SO_4 \to MgSO_4 + H-2$
b) $n_{H_2SO_4} = n_{Mg} = \dfrac{4,8}{24} = 0,2(mol)$
$C\%_{H_2SO_4} = \dfrac{0,2.98}{200}.100\% = 9,8\%$
$n_{H_2} = n_{Mg} = 0,2(mol)$
$\Rightarrow m_{dd\ A} = 4,8 + 200 - 0,2.2 = 204,4(gam)$
$C\%_{MgSO_4} = \dfrac{0,2.120}{204,4}.100\% = 11,7\%$
c) $V_{H_2} = 0,2.22,4 = 4,48(lít)$
nHCl= (7,3%.250)/100=0,5(mol)
nCuO=0,05(mol)
a) PTHH: CuO +2 HCl -> CuCl2 + H2O
Ta có: 0,5/2 > 0,05/1
=> HCl dư, CuO hết => tính theo nCuO
b) nCuCl2=nCuO=0,05(mol) => mCuCl2= 135. 0,05= 6,75(g)
c) nHCl(dư)=0,5-0,05.2=0,4(mol) => mHCl(dư)=0,4.36,5=14,6(g)
mddsau=250+4= 254(g)
=>C%ddCuCl2= (6,75/254).100=2,657%
C%ddHCl(dư)= (14,6/254).100=5,748%
\(n_{CuSO_4}=1\cdot0,3=0,3\left(mol\right);n_{NaOH}=2\cdot0,07=0,14\left(mol\right)\\ PTHH:CuSO_4+2NaOH\rightarrow Na_2SO_4+Cu\left(OH\right)_2\downarrow\\ \text{Vì }\dfrac{n_{CuSO_4}}{1}>\dfrac{n_{NaOH}}{2}\Rightarrow CuSO_4\text{ dư}\\ \Rightarrow n_{Na_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,07\left(mol\right)\\ \Rightarrow C_{M_{Na_2SO_4}}=\dfrac{0,07}{0,3+0,07}\approx1,19M\)
nZn(OH)2= 29,7/99=0,3(mol)
mHCl= 20%. 164,25=32,85(g) => nHCl= 32,85/36,5=0,9(mol)
PTHH: Zn(OH)2 +2 HCl -> ZnCl2 + H2O
Ta có: 0,9/2 > 0,3/1
=> Zn(OH)2 hết, HCl dư => tính theo nZn(OH)2
Ta có: nZnCl2 = nZn(OH)2=0,3(mol) => mZnCl2= 0,3.136= 40,8(g)
nHCl(dư)=0,9 - 0,3.2=0,3(mol) => mHCl(dư)=0,3.36,5=10,95(g)
mddsau= mZn(OH)2 + mddHCl = 29,7+ 164,25=193,95(g)
=> C%ddHCl(dư)= (10,95/193,95).100=5,646%
C%ddZnCl2= (40,8/193,95).100=21,036%
a, PT: \(Zn+CuSO_4\rightarrow ZnSO_4+Cu_{\downarrow}\)
b, Ta có: \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
\(m_{CuSO_4}=\dfrac{250.20}{100}=50\left(g\right)\Rightarrow n_{CuSO_4}=\dfrac{50}{160}=0,3125\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,3125}{1}\) , ta được CuSO4 dư.
Kim loại thu được sau pư là Cu.
Theo PT: \(n_{Cu}=n_{Zn}=0,2\left(mol\right)\)
\(\Rightarrow m_{Cu}=0,2.64=12,8\left(g\right)\)
b, Theo PT: \(n_{ZnSO_4}=n_{CuSO_4\left(pư\right)}=n_{Zn}=0,2\left(mol\right)\)
\(\Rightarrow n_{CuSO_4\left(dư\right)}=0,1125\left(mol\right)\)
Dung dịch thu được sau pư gồm: ZnSO4 và CuSO4 dư.
Ta có: m dd sau pư = mZn + m dd CuSO4 - mCu = 13 + 250 - 12,8 = 250,2 (g)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{ZnSO_4}=\dfrac{0,2.161}{250,2}.100\%\approx12,87\%\\C\%_{CuSO_4\left(dư\right)}=\dfrac{0,1125.160}{250,2}.100\%\approx7,19\%\end{matrix}\right.\)
Bạn tham khảo nhé!