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\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
PTHH: Zn + H2SO4 ---> ZnSO4 + H2
0,2--->0,2--------->0,2------>0,2
\(V_{H_2}=0,2.22,4=4,48\left(l\right)\\ m_{H_2SO_4}=\dfrac{0,2.98}{20\%}=98\left(g\right)\\ \rightarrow V_{ddH_2SO_4}=\dfrac{98}{1,14}=86\left(ml\right)=0,086\left(l\right)\\ \rightarrow C_{M\left(H_2SO_4\right)}=\dfrac{0,2}{0,086}=2,33M\)
`Zn+H_2SO_4->ZnSO_4+H_2`(to)
0,45-------------------0,45------0,45mol
`n_(Zn)=(29,25)/65=0,45mol`
`m_(ZnSO_4)=0,45.161=72,45g`
`V_(H_2)=0,45.22,4=10,08l`
c) `H_2+CuO->Cu+H_2O`(to)
0,45--------0,45 mol
`n_(Cu)=40/80=0,5 mol`
=>Cu dư , 0,05 mol
`m_(chất rắn)=0,45.64+0,05.80=32,8g`
\(n_{Zn}=\dfrac{m}{M}=\dfrac{29,25}{65}=0,45\left(mol\right)\)
a) \(PTHH:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
1 1 1 1
0,45 0,45 0,45 0,45
b) \(m_{ZnSO_4}=n.M=0,45.\left(65+32+16.4\right)=51,03\left(g\right)\\ V_{H_2}=n.24,79=0,45.24,79=11,1555\left(l\right)\)
c) \(n_{CuO}=\dfrac{m}{M}=\dfrac{40}{\left(64+16\right)}=0,5\left(mol\right)\)
\(PTHH:CuO+H_2\rightarrow Cu+H_2O\)
1 1 1 1
0,5 0,5 0,5 0,5
\(m_{Cu}=0,5.64=32\left(g\right).\)
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
\(PTHH:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
(mol)_____0,2____0,2______0,2____0,2__
\(a.V_{H_2}=22,4.0,2=4,48\left(l\right)\)
\(b.m_{ddH_2SO_4}=\dfrac{0,2.98.100}{24,5}=80\left(g\right)\)
\(c.m_{ddspu}=13+80-0,2.2=92,6\left(g\right)\\ \Rightarrow C\%_{ddspu}=\dfrac{0,2.136}{92,6}.100=29,4\left(\%\right)\)
nHCl = 0.3*0.75 = 0.225 (mol)
Zn + 2HCl => ZnCl2 + H2
0.1125..0.225....0.1125..0.1125
mZn = 0.1125*65 = 7.3125 (g)
VH2 = 0.1125*22.4 = 2.52 (l)
CM ZnCl2 = 0.1125/0.3 = 0.375 (M)
nHCl= 0,75 x 0,3= 0,225(mol)
PTHH: Zn + 2HCl -> ZnCl2 + H2
a) 0,1125____0,225___0,1125___0,1125(mol)
=> a=mZn=0,1125.65=7,3125(g)
V(H2,đktc)=0,1125 x 22,4=2,52(l)
b) VddZnCl2=VddHCl=0,3(l)
=>CMddZnCl2=0,1125/0,3=0,375(M)
\(a,n_{Zn}=\dfrac{1,3}{65}=0,02\left(mol\right)\)
PTHH: Zn + H2SO4 ---> ZnSO4 + H2
0,02--->0,02--------->0,02----->0,02
b, mZnSO4 = 0,02.161 = 3,22 (g)
c, VH2 = 0,02.22,4 = 0,448 (l)
d, \(m_{ddH_2SO_4}=\dfrac{0,02.98}{10\%}=19,6\left(g\right)\)
e, mdd = 19,6 + 1,3 - 0,02.2 = 20,86 (g)
=> \(C\%_{ZnSO_4}=\dfrac{0,02.161}{20,86}.100\%=15,44\%\)
\(n_{Fe}=\dfrac{22,4}{56}=0,4\) (mol) (1)
Phương trình hóa học :
Fe + 2HCl ---> FeCl2 + H2 (2)
Từ (1) và (2) ta có \(n_{FeCl_2}=n_{H_2}=0,4\) (mol) ; \(n_{HCl}=0,8\left(mol\right)\)
b) => \(m_{\text{muối}}=0,4.\left(56+35,5.2\right)=50.8\left(g\right)\)
c) \(V_{\text{khí}}=0,4.22,4=8,96\left(l\right)\)
d) \(m_{HCl}=0,8.36.5=29,2\left(g\right)\)
\(\Rightarrow C\%=\dfrac{29,2}{200}.100\%=14,6\%\)
a) Pt: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
b) nFe = \(\dfrac{11,2}{56}=0,2mol\)
Theo pt: nH2 = nFe = 0,2 mol
=> VH2 = 0,2.22,4 = 4,48lit
c) Theo pt: nHCl = 2nFe = 0,4 mol
=> mHCl = 0,4.36,5 = 14,6 g
=> C% = \(\dfrac{14,6}{73}.100\%=20\%\)
a) Pt:
b) nFe = \(\dfrac{11,2}{56}=0,2mol\)
Theo pt: nH2 = nFe = 0,2 mol
=> VH2 = 0,2.22,4 = 4,48lit
c) Theo pt: nHCl = 2nFe = 0,4 mol
=> mHCl = 0,4.36,5 = 14,6 g
=> \(C\%=\dfrac{14,6}{73}.100\%=20\%\)
a)
\(Zn + 2HCl \to ZnCl_2 + H_2\)
b),c)
Theo PTHH :
\(n_{ZnCl_2} = n_{H_2} = n_{Zn} = \dfrac{13}{65} = 0,2(mol)\)
Vậy :
\(m_{ZnCl_2} = 0,2.136 = 27,2(gam)\\ V_{H_2} =0,2.22,4 = 4,48(lít)\)
a) PTHH: 4Zn + 5H2SO4 -> 4ZnSO4 + H2S + 4H2O
a) nZn=m/M=1,3/65=0,02(mol)
PT: Zn + H2SO4 -> ZnSO4 + H2
1...........1...............1................1 (mol)
0,02........0,02.........0,02.............0,02 (mol)
VH2=n.22,4=0,02.22,4=0,448(lít
mH2SO4=n.M=0,02.(2+32+64)=1,96(g)
=> \(C\%_{H_2SO_4}=\dfrac{m_{H_2SO_4}.100\%}{m_{ddH_2SO_4}}=\dfrac{1,96.100\%}{98}=2\left(\%\right)\)