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nZn = 13/65 = 0.2 (mol)
Zn + H2SO4 => ZnSO4 + H2
0.2......0.2..........................0.2
VH2 = 0.2*22.4 = 4.48 (l)
C%H2SO4 = 0.2*98/200 * 100% = 9.8 %
nCuO = 8/80 = 0.1 (mol)
CuO + H2 -to-> Cu + H2O
0.1......0.1...........0.1
=> H2 dư
mCu = 0.1*64 = 6.4 (g)
â) nZn=0,2(mol)
PTHH: Zn + H2SO4 -> ZnSO4 + H2
0,2_____0,2______0,2_____0,2(mol)
=> V(H2,đktc)=0,2.22,4=4,48(l)
b) C%ddH2SO4= [(98.0,2)/200)].100=9,8%
c) nCuO=0,1(mol)
PTHH: CuO + H2 -to-> Cu + H2O
Ta có: 0,1/1 < 0,2/1
=> H2 dư, CuO hết, tính theo nCuO
=> nCu=nCuO=0,1(mol)
=>mCu=6,4(g)
$a\big)2Al+6HCl\to 2AlCl_3+3H_2$
$b\big)$
$n_{Al}=\dfrac{5,4}{27}=0,2(mol)$
Theo PT: $n_{H_2}=\dfrac{3}{2}n_{Al}=0,3(mol)$
$\to V_{H_2(đktc)}=0,3.22,4=6,72(l)$
$c\big)$
Theo PT: $n_{AlCl_3}=n_{Al}=0,2(mol)$
$\to m_{AlCl_3}=0,2.133,5=26,7(g)$
1:
a) \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
PTHH: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\)
______0,2------>0,2------------------->0,2_____(mol)
=> \(V_{H_2}=0,2.22,4=4,48\left(l\right)\)
b) \(V_{ddH_2SO_4}=\dfrac{0,2}{1}=0,2\left(l\right)\)
2:
a)
\(n_{HCl}=2.0,2=0,4\left(mol\right)\)
PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
______0,2<------0,4------------------>0,2______(mol)
=> \(m_{Mg}=0,2.24=4,8\left(g\right)\)
b) \(V_{H_2}=0,2.22,4=4,48\left(l\right)\)
a.b.\(n_{Al}=\dfrac{m_{Al}}{M_{Al}}=\dfrac{5,4}{27}=0,2mol\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
0,2 0,3 ( mol )
\(V_{H_2}=n_{H_2}.22,4=0,3.22,4=6,72l\)
c.\(n_{CuO}=\dfrac{m_{CuO}}{M_{CuO}}=\dfrac{32}{80}=0,4mol\)
\(CuO+H_2\rightarrow Cu+H_2O\)
0,4 < 0,3 ( mol )
0,3 0,3 0,3 ( mol )
\(m_A=m_{CuO\left(du\right)}+m_{Cu}=\left[\left(0,4-0,3\right).80\right]+\left(0,3.64\right)=8+19,2=27,2g\)
a) $Fe + 2HCl \to FeCl_2 + H_2$
Theo PTHH : n H2 = n Fe = 8,4/56 = 0,15(mol)
V H2 = 0,15.22,4 = 3,36(lít)
b) n HCl = 2n Fe = 0,3(mol)
=> CM HCl = 0,3/0,2 = 1,5M
c) $CuO + H_2 \xrightarrow{t^o} Cu + H_2O$
Ta thấy :
n CuO = 32/80 = 0,4 > n H2 = 0,15 mol nên CuO dư
Theo PTHH : n Cu = n H2 = 0,15 mol
=> m Cu = 0,15.64 = 9,6 gam
\(n_{Al}=\dfrac{16,2}{27}=0,6\left(mol\right)\)
2Al + 3H2SO4 ---> Al2(SO4)3 + 3H2
0,6 0,9 0,3 0,9
\(\rightarrow V_{H_2}=0,9.22,4=20,16\left(l\right)\)
\(n_{Cu}=\dfrac{57}{64}=0,890625\left(mol\right)\)
PTHH: CuO + H2 --to--> Cu + H2O
0,890625 0,890625
\(H=\dfrac{0,890625}{0,9}=99\%\)
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