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a)
$n_{Zn} = \dfrac{13}{65} = 0,2(mol) ; n_{H_2 SO_4} = 0,5.2 = 1(mol)$
$Zn + H_2SO_4 \to ZnSO_4 + H_2$
Ta thấy :
$n_{Zn} < n_{H_2SO_4}$ nên $H_2SO_4$ dư
$n_{ZnSO_4} = n_{H_2SO_4\ pư} = n_{Zn} = 0,2(mol)$
$m_{ZnSO_4} = 0,2.161=32,2(gam)$
$m_{H_2SO_4\ pư} = 0,2.98 = 19,6(gam)$
b)
$n_{H_2SO_4\ dư} = 1 - 0,2 = 0,8(mol)$
$C_{M_{H_2SO_4\ dư}} = \dfrac{0,8}{0,5} = 1,6M$
$C_{M_{FeSO_4}} = \dfrac{0,2}{0,5} = 0,4M$
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ n_{H_2SO_4}=0,5.2=1\left(mol\right)\\ Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ a.Vì:\dfrac{0,2}{1}< \dfrac{1}{1}\Rightarrow H_2SO_4dư\\ n_{H_2SO_4\left(p.ứ\right)}=n_{ZnSO_4}=n_{Zn}=0,2\left(mol\right)\\ m_{H_2SO_4\left(p.ứ\right)}=0,2.98=19,6\left(g\right)\\ m_{ZnSO_4}=161.0,2=32,2\left(g\right)\\ b.V_{ddsau}=V_{ddH_2SO_4}=0,5\left(l\right)\\ C_{MddZnSO_4}=\dfrac{0,2}{0,5}=0,4\left(M\right)\\ C_{MddH_2SO_4\left(dư\right)}=\dfrac{1-0,2}{0,5}=1,6\left(M\right)\)
\(n_{Zn}=\dfrac{26}{65}=0,4\left(mol\right)\)
Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2|\)
1 2 1 1
0,4 0,8 0,4 0,4
a) \(n_{H2}=\dfrac{0,4.1}{1}=0,4\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,4.22,4=8,96\left(l\right)\)
b) \(n_{HCl}=\dfrac{0,4.2}{1}=0,8\left(mol\right)\)
⇒ \(m_{HCl}=0,8.36,5=29,2\left(g\right)\)
\(m_{ddHCl}=\dfrac{29,2.100}{14,6}=200\left(g\right)\)
c) \(n_{ZnCl2}=\dfrac{0,4.1}{1}=0,4\left(mol\right)\)
⇒ \(m_{ZnCl2}=0,4.136=54,4\left(g\right)\)
\(m_{ddspu}=26+200-\left(0,4.2\right)=225,2\left(g\right)\)
\(C_{ZnCl2}=\dfrac{54,4.100}{225,2}=24,16\)0/0
Chúc bạn học tốt
\(n_{Zn}=\dfrac{13}{65}=0,2mol\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,2 0,4 0,2 0,2
a)\(m_{HCl}=0,4\cdot36,5=14,6g\)
\(m_{ddHCl}=\dfrac{14,6}{18,25\%}\cdot100\%=80g\)
b)\(V_{H_2}=0,2\cdot22,4=4,48l\)
c)\(m_{H_2}=0,2\cdot2=0,4g\)
BTKL: \(m_{Zn}+m_{ddHCl}=m_{ddZnCl_2}+m_{H_2}\)
\(\Rightarrow m_{ddZnCl_2}=13+80-0,4=92,6g\)
\(m_{ctZnCl_2}=0,2\cdot136=27,2g\)
\(C\%=\dfrac{27,2}{92,6}\cdot100\%=29,37\%\)
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1mol\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,1 0,2 0,1 ( mol )
\(m_{Zn}=0,1.65=6,5g\)
\(C_{M_{HCl}}=\dfrac{0,2}{0,25}=0,8M\)
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\
pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,1 0,2 0,1
\(m_{Zn}=0,1.65=6,5g\\
C_{M\left(HCl\right)}=\dfrac{0,2}{0,25}=0,8M\)
a, \(Mg+2HCl\rightarrow MgCl_2+H_2\)
b, \(m_{Mg}=\dfrac{7,2}{24}=0,3\left(mol\right)\)
Theo PT: \(n_{MgCl_2}=n_{Mg}=0,3\left(mol\right)\Rightarrow m_{MgCl_2}=0,3.95=28,5\left(g\right)\)
c, \(n_{HCl}=2n_{Mg}=0,6\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}}=\dfrac{0,6}{0,2}=3\left(M\right)\)
Nzn=0,1mol
a) có pt : Zn + 2Hcl -> ZnCl2 + H2
1 -> 2 -> 1 -> 1 mol
0,1-> 0,2 -> 0,1 -> o,1 mol
b) số Hcl đã dùng khi pứ là :
0,2 . 36,5= 7,3
-> số gam hcl dư là 10,95- 7,3=3,65(g)
c)mZnCl2=0,1 . 136=13,6g
ZnO+2HNO3------>Zn(NO3)2 + H2O
a) Ta có
n\(_{ZnO}=\frac{16,2}{81}=0,2\left(mol\right)\)
m\(_{HNO3}=\frac{200.15}{100}=30\left(g\right)\)
n\(_{HNO3}=\frac{30}{63}=0,48\left(mol\right)\)
=> HNO3 dư
Theo ptthh
n\(_{HNO3}=2n_{ZnO}=0,4\left(mol\right)\)
m\(_{HNO3}=63.0,4=25,2\left(g\right)\)
b) Theo ptth
n\(_{Zn\left(NO3\right)2}=n_{ZnO}=0,2\left(mol\right)\)
M\(_{Zn\left(NO3\right)2}=0,2.189=37,9\left(mol\right)\)
c)m dd=400+16,2=416,2(g)
m HNO3 dư=( 0,48-0,4).63=5,04(g)
C% (HNO3)=\(\frac{5,04}{416,2}.100\%=1,21\%\)
C% (Zn(NO3)2 =\(\frac{37,9}{416,2}.100\%=9,1\%\)
Chúc bạn học tốt
nZnO= 16,2/81=0,1 mol
mHNO3= \(\frac{400.15}{100}60\left(g\right)\)
PTHH:\(Zn+2HNO3\rightarrow Zn\left(NO3\right)+H2O\)
Theo TLPT : HNO3 dư
nHNO3 dư = 0,9 - 0,4 = 0,5
a, Theo PT: nHNO3 = 0,4 (mol)
=>mHNO3 = 0,4.63= 25,2
b, Theo PT: nZn(NO3)2 = 0,2 (mol)
=>mZN(NO3)2= 0,2.230=46(g)
c, Các chất còn lại có trong dd sau pư HNO3 (dư)
m HNO3 dư = 0,5.0,3= 34,65(g)
m dd sau pư : 16,24+400=416,2(g)
C%Zn(NO3)2= 11,05 %
C% HNO3 dư = 8,33%