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\(4.a/n_{Al}=\dfrac{5,4}{27}=0,2mol\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
0,2 0,3 0,1 0,3
\(V_{H_2}=0,3.24,79=7,437l\\ b/C_{\%H_2SO_4}=\dfrac{0,3.98}{150}\cdot100=19,6\%\\ c/m_{Al_2\left(SO_4\right)_3}=0,1.342=34,2g\)
\(5.a/n_{MgO}=\dfrac{4}{40}=0,1mol\\ MgO+2HCl\rightarrow MgCl_2+H_2O\)
0,1 0,2 0,1 0,1
\(C_{\%HCl}=\dfrac{0,2.36,5}{200}\cdot100=3,65\%\\ b/C_{\%MgCl_2}=\dfrac{0,1.95}{200+4}\cdot100=4,66\%\\ c/NaOH+HCl\rightarrow NaCl+H_2O\\ n_{NaOH}=n_{HCl}=0,2mol\\ V_{NaOH}=\dfrac{0,2}{1}=0,2l=200ml\)
\(4.\\ a/n_{Al}=\dfrac{5,4}{27}=0,2mol\\2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(0,2......0,3............0,1...............0,3\)
\(V_{H_2}=0,3.24,79=7,437l\\ b.C_{\%H_2SO_4}=\dfrac{0,3.98}{150}\cdot100=19,6\%\\ c.m_{Al_2\left(SO_4\right)_3}=0,1.342=34,2g\)
\(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)
PTHH :
\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\)
0,1 0,1 0,1 0,1
\(a,m_{MgSO_4}=0,1.120=12\left(g\right)\)
\(b,V_{H_2}=0,1.22,4=2,24\left(l\right)\)
\(c,m_{ddH_2SO_4}=\dfrac{0,1.98.100}{10}=98\left(g\right)\)
PT: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
Ta có: \(n_{H_2}=\dfrac{3,7185}{24,79}=0,15\left(mol\right)\)
a, \(n_{Al}=\dfrac{2}{3}n_{H_2}=0,1\left(mol\right)\Rightarrow m_{Al}=0,1.27=2,7\left(g\right)\)
b, \(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{3}n_{H_2}=0,05\left(mol\right)\)
\(\Rightarrow m_{Al_2\left(SO_4\right)_3}=0,05.342=17,1\left(g\right)\)
c, \(n_{H_2SO_4}=n_{H_2}=0,15\left(mol\right)\)
\(\Rightarrow V_{H_2SO_4}=\dfrac{0,15}{1,5}=0,1\left(l\right)=100\left(ml\right)\)
\(n_{Al}=\dfrac{2,7}{27}=0,1mol\\ a.2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
0,1 0,15 0,05 0,15
\(b.V_{H_2}=0,15.24,79=3,7185l\\ c.m_{Al_2\left(SO_4\right)_3}=0,05.342=17,1g\\ d.C_{M_{H_2SO_4}}=\dfrac{0,15}{0,4}=0,375M\)
a) 2Al+6HCl→→2AlCl3+3H2
b)
nAl=10,8\27=0,4(mol)
nAlCl3=nAl=0,4(mol)
mAlCl3=0,4.133,5=53,4(g)
c)
nH2=3\2nAl=0,6(mol)
VH2=22,4.0,6=13,44(l)
d) n HCl=0,4.6\2=1,2 mol
=>Cm HCl=1,2\0,1=12M
\(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)
a) Pt : \(2Al+6HCl\rightarrow2AlCl_3+3H_2|\)
2 6 2 3
0,4 1,2 0,4 0,6
b) \(n_{H2}=\dfrac{0,4.3}{2}=0,6\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,6.24,79=14,874\left(l\right)\)
c) \(n_{AlCl3}=\dfrac{0,6.2}{3}=0,4\left(mol\right)\)
⇒ \(m_{AlCl3}=0,4.133,5=53,4\left(g\right)\)
d) \(n_{HCl}=\dfrac{0,4.6}{2}=1,2\left(mol\right)\)
100ml = 0,1l
\(C_{M_{ddHCl}}=\dfrac{1,2}{0,1}=12\left(M\right)\)
Chúc bạn học tốt
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)
\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ 0,2........0,3...........0,1...........0,3\left(mol\right)\\ a.C_{MddH_2SO_4}=\dfrac{0,3}{0,3}=1\left(M\right)\\ b.m_{Al_2\left(SO_4\right)_3}=342.0,1=34,2\left(g\right)\\ c.V_{H_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\)
PTHH: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
a) \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{3}{2}n_{Al}=0,3\left(mol\right)\)
\(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,3}{0,3}=1M\)
b) \(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Al}=0,1\left(mol\right)\)
\(\Rightarrow m_{Al_2\left(SO_4\right)_3}=n.M=0,1.342=34,2\left(g\right)\)
c) \(n_{H_2}=n_{H_2SO_4}=0,3\left(mol\right)\)
\(\Rightarrow V_{H_2}=n.22,4=0,3.22,4=6,72\left(l\right)\)