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a) Mg + H2SO4 --> MgSO4 + H2
b) \(n_{Mg}=\dfrac{14,4}{24}=0,6\left(mol\right)\)
PTHH: Mg + H2SO4 --> MgSO4 + H2
0,6--->0,6------->0,6----->0,6
=> \(m_{H_2SO_4}=0,6.98=58,8\left(g\right)\)
c)
PTHH: 2H2 + O2 --to--> 2H2O
0,6-->0,3
=> VO2 = 0,3.24,79 = 7,437 (l)
=> Vkk = 7,437.5 = 37,185 (l)
\(n_{Al}=\dfrac{2,7}{27}=0,1mol\\ a.2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
0,1 0,15 0,05 0,15
\(b.V_{H_2}=0,15.24,79=3,7185l\\ c.m_{Al_2\left(SO_4\right)_3}=0,05.342=17,1g\\ d.C_{M_{H_2SO_4}}=\dfrac{0,15}{0,4}=0,375M\)
2Al + 3H2SO4 -- > Al2(SO4)3 + 3H2
0,3 0,45 0,15 0,45
nAl = 8,1 / 27 = 0,3(mol)
\(VH_2=0,45.22,4=10,08\left(g\right)\)
\(m\left(muối\right)=0,15.342=51,3\left(g\right)\)
\(H_2+CuO\rightarrow Cu+H_2O\)
0,45 0,45
mCu = 0,45 . 64 = 28,8(g)
bạn giải thích dùm mình tại sao 3H2So4 với 3H2 lại là 0,45 mol ko
\(n_{Zn}=\dfrac{9,75}{65}=0,15\left(mol\right)\\
pthh:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
0,15 0,15 0,15 0,15
\(V_{H_2}=0,15.22,4=3,36L\\
m_{H_2SO_4}=0,15.98=14,7\left(g\right)\\
m_{ZnSO_4}=161.0,15=24,15g\\
\)
\(n_{CuO}=\dfrac{6}{80}=0,075\left(mol\right)\\
pthh:CuO+H_2\underrightarrow{t^o}Cu+H_2O\\
LTL:0,075< 0,15\)
=> H2 dư
\(n_{Cu}=n_{CuO}=0,075\left(mol\right)\\
m_{Cu}=0,075.64=4,8g\)
nFe = \(\dfrac{14}{56}=0,25\) mol
Fe + H2SO4 → FeSO4 + H2
0,25 → 0,25 → 0,25 → 0,25
mH2SO4 = 0,25.98 = 24,5 gam
mFeSO4 = 0,25.152 = 38 gam
VH2 = 0,25.22,4 = 5,6 gam
nFe = 1456=0,251456=0,25 mol
Fe + H2SO4 → FeSO4 + H2
0,25 → 0,25 → 0,25 → 0,25
mH2SO4 = 0,25.98 = 24,5 gam
mFeSO4 = 0,25.152 = 38 gam
VH2 = 0,25.22,4 = 5,6 gam.
Bài 1 : Sửa ZnSO thành ZnSO4
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
Pt : \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
Theo Pt : \(n_{Zn}=n_{H2SO4}=n_{ZnSO4}=n_{H2}=0,2\left(mol\right)\)
a) \(m_{H2SO4}=0,2.98=19,6\left(g\right)\)
b) \(m_{ZnSO4}=0,2.161=32,2\left(g\right)\)
c) \(V_{H2\left(dkc\right)}=0,2.24,79=4,958\left(l\right)\)
Bài 3 :
\(n_{O2}=\dfrac{9,6}{32}=0,3\left(mol\right)\)
\(2KClO_3\xrightarrow[]{t^o}2KCl+3O_2\)
0,2<-----------0,2<----0,3
a) \(m_{KClO3}=0,2.122,5=24,5\left(g\right)\)
b) Cách 1 : \(m_{KCl}=0,2.74,5=14,9\left(g\right)\)
cách 2 : \(BTKl:m_{KClO3}=m_{KCl}+m_{O2}\)
\(\Rightarrow m_{KCl}=m_{KClO3}-m_{O2}=24,5-9,6=14,9\left(g\right)\)
PT: \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
Ta có: \(n_{CH_4}=\dfrac{30,9875}{24,79}=1,25\left(mol\right)\)
a, \(n_{H_2O}=2n_{CH_4}=2,5\left(mol\right)\) \(\Rightarrow m_{H_2O}=2,5.18=45\left(g\right)\)
b, \(n_{O_2}=2n_{CH_4}=2,5\left(mol\right)\) \(\Rightarrow V_{O_2}=2,5.24,79=61,975\left(l\right)\)
Mà: O2 chiếm 1/5 thể tích không khí.
\(\Rightarrow V_{kk}=5V_{O_2}=309,875\left(l\right)\)
\(n_{Al}=\dfrac{m}{M}=\dfrac{0,54}{27}=0,02mol\)
\(n_{H_2SO_4}=\dfrac{m}{M}=\dfrac{22,05}{98}=0,225mol\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
0,02 < 0,225 ( mol )
0,02 0,03 ( mol )
\(V_{H_2}=n.22,4=0,03.24,79=0,7437l\)
\(n_{H_2}=\dfrac{4,958}{24,79}=0,2\left(mol\right)\)
PTHH: 2Al + 3H2SO4 ---> Al2(SO4)3 + 3H2
2/15<-------------------1/15<----------0,2
2H2 + O2 --to--> 2H2O
0,2-->0,1
\(m_{Al}=\dfrac{2}{15}.27=3,6\left(g\right)\\ m_{Al_2\left(SO_4\right)_3}=\dfrac{1}{15}.342=22,8\left(g\right)\\ V_{kk}=5.0,1.24,79=12,395\left(mol\right)\)