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1/ Fe +2 HCl --------> FeCl2 + H2
\(n_{Fe}=\dfrac{2,8}{56}=0,05\left(mol\right)\)
\(n_{H_2}=n_{Fe}=0,05\left(mol\right)\Rightarrow V_{H_2}=0,05.22,4=1,12\left(l\right)\)
\(n_{HCl}=2n_{Fe}=0,1\left(mol\right)\Rightarrow m_{HCl}=0,1.36,5=3,65\left(g\right)\)
Câu 6 :
1) $n_{Fe} = \dfrac{2,8}{56} = 0,05(mol)$
Fe + 2HCl → FeCl2 + H2
0,05...0,1....................0,05......(mol)$
$V_{H_2} = 0,05.22,4 = 1,12(lít)$
$m_{HCl} = 0,1.36,5 = 3,65(gam)$
2)
a) $CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O$
$V_{O_2} = 2V_{CH_4} = 4(lít)$
b) $n_{CO_2} = n_{CH_4} = 0,15(mol) \Rightarrow V_{CO_2} = 0,15.22,4 = 3,36(lít)$
c) $d_{CH_4/kk} = \dfrac{16}{29} = 0,552$
Vậy khí metan nhẹ hơn không khí 0,552 lần
Zn+2H2SO4->ZnSO4+H2
0,2----------------------------0,2
n Zn=\(\dfrac{13}{65}\)=0,2 mol
=>VH2=0,2.22,4=4,48l
b)
Fe+2HCl->FeCl2+H2
0,1--------------0,05 mol
=>VH2=0,05.22,4=1,12l
\(a,n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ n_{H_2}=n_{Zn}=0,2\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\\ b,Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{HCl}=0,1\left(mol\right)\Rightarrow n_{H_2}=\dfrac{0,1}{2}=0,05\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,05.22,4=1,12\left(l\right)\)
nHCl=5.10-3 mol
2Na + 2H2O --> 2NaOH + H2
x mol x mol 1/2 mol
Ba + 2H2O --> Ba(OH)2 + H2
y mol y mol y mol
NaOH + HCl --> NaCl + H2O
x mol x mol
Ba(OH)2 + 2HCl--> BaCl2 + H2O
y mol 2y mol
Ta duoc: 23x + 137y =0,297 (1)
x + 2y =5.10-3 (2)
Tu (1) va (2) ta duoc => x= 10-3
=> y= 2.10-3
a/ mNa= 10-3.23=0,023g
mBa=2.10-3.137=0,274g
b/ nH2= 10-6 mol
H2 + O2 --> H2O
10-6 mol 10-6 mol
VO2= 10-6. 22,4=2,24.10-5 lit
VKK= 2,24.10-5.100/20=1,12.10-4 lit
\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
a. \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
Theo PTHH: \(n_{ZnSO_4}=n_{Zn}=0,1\left(mol\right)\)
\(\Rightarrow m_{muối}=0,1.161=16,1\left(g\right)\)
b. \(n_{H_2thu.được}=n_{Zn}=0,1\left(mol\right)\)
\(H_2+\dfrac{1}{2}O_2\underrightarrow{t^o}H_2O\)
0,1 0,05
\(V_{O_2}=0,05.22,4=1,12\left(l\right)\)
\(\Rightarrow V_{không.khí}=1,12.5=5,6\left(l\right)\)
\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{49}{98}=0,5\left(mol\right)\)
PTHH: 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
Xét tỉ lệ: \(\dfrac{0,2}{2}< \dfrac{0,5}{3}\) => Al hết, H2SO4 dư
PTHH: 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
0,2---------------------------->0,3
=> VH2 = 0,3.22,4 = 6,72 (l)
\(n_{Al}=\dfrac{m}{M}=\dfrac{5,4}{27}=0,2=\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{m}{M}=\dfrac{49}{98}=0,5\left(mol\right)\)
2Al + 3H2SO4 → Al2(SO4)3 + 3H2
2 3 ( mol )
0,2 0,5 ( mol )
Tỉ lệ: \(\dfrac{0,2}{2}< \dfrac{0,5}{3}\) ⇒ H2SO4 dư
2Al + 3H2SO4 → Al2(SO4)3 + 3H2
0,2 → 0,3 → 0,3 ( mol )\(V_{H_2}=n.22,4=0,3.22,4=6,72\left(l\right)\)
\(a,n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ n_{HCl}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\)
PTHH: Mg + 2HCl ---> MgCl2 + H2
LTL: \(0,2< \dfrac{0,5}{2}\) => HCl dư
Theo pthh: nH2 = nMg = 0,2 (mol)
=> VH2 = 0,2.22,4 = 4,48 (l)
\(b,n_{Fe}=\dfrac{2,8}{56}=0,.05\left(mol\right)\\ n_{H_2SO_4}=\dfrac{9,8}{98}=0,1\left(mol\right)\)
PTHH: Fe + H2SO4 ---> FeSO4 + H2
LTL: 0,05 < 0,1 => H2SO4 dư
Theo pthh: nH2 = nFe = 0,05 (mol)
=> VH2 = 0,05.22,4 = 1,12 (l)
\(c,n_{Zn}=\dfrac{14,95}{65}=0,23\left(mol\right)\\ n_{HCl}=\dfrac{21,9}{36,5}=0,6\left(mol\right)\)
PTHH: Zn + 2HCl ---> ZnCl2 + H2
LTL: \(0,23< \dfrac{0,6}{2}\) => HCl dư
Theo pthh: nH2 = nZn = 0,23 (mol)
=> VH2 = 0,23.22,4 = 5,152 (l)
- PTHH : \(2NaHCO_3+H_2SO_4\rightarrow Na_2SO_4+2H_2O+2CO_2\)
\(n_{H_2SO_4}=\frac{m}{M}=\frac{980}{1.2+32+16.4}=\frac{980}{98}=10\left(mol\right)\)
- Theo PTHH : \(n_{H_2SO_4}=2n_{CO_2}=2.10=20\left(mol\right)\)
-> \(V_{CO_2}=n_{CO_2}.22,4=20.22,4=448\left(l\right)\)