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\(a,n_{Mg}=\dfrac{7,2}{24}=0,3\left(mol\right)\\ PTHH:Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
\(b,\) Theo PT: \(n_{H_2}=n_{Mg}=0,3\left(mol\right)\)
\(\Rightarrow V_{H_2\left(đktc\right)}=0,3\cdot22,4=6,72\left(l\right)\)
\(c,n_{H_2SO_4}=n_{Mg}=0,3\left(mol\right)\\ \Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,3}{0,2}=1,5M\)
\(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)
PTHH :
\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\)
0,1 0,1 0,1 0,1
\(a,m_{MgSO_4}=0,1.120=12\left(g\right)\)
\(b,V_{H_2}=0,1.22,4=2,24\left(l\right)\)
\(c,m_{ddH_2SO_4}=\dfrac{0,1.98.100}{10}=98\left(g\right)\)
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ Mg+H_2SO_4\rightarrow MgSO_4+H_2\\ n_{MgSO_4}=n_{H_2}=n_{H_2SO_4}=n_{Mg}=0,2\left(mol\right)\\ a,m_{MgSO_4}=120.0,2=24\left(g\right)\\ b,V_{H_2\left(đkc\right)}=24,79.0,2=4,958\left(l\right)\\ c,Oxide:A_2O_x\left(x:hoá.trị.A\right)\\ A_2O_x+xH_2SO_4\rightarrow A_2\left(SO_4\right)_x+xH_2O\\ n_{Oxide}=\dfrac{\dfrac{3}{4}.0,2.1}{x}=\dfrac{0,15}{x}\left(mol\right)\\ M_{A_2O_x}=\dfrac{8}{\dfrac{0,15}{x}}=\dfrac{160}{3}x\)
Xét x=1;x=2;x=3;x=8/3 thấy x=3 (TM) khi đó KLR oxide là 160g/mol
\(M_{M_2O_3}=2M_M+3.16=160\left(\dfrac{g}{mol}\right)\\ \Rightarrow M_M=\dfrac{160-48}{2}=56\left(\dfrac{g}{mol}\right)\)
Nên: M là sắt (Fe=56)
Oxide CTHH: Fe2O3
Ta có: \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
PT: \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
Theo PT: \(n_{H_2SO_4}=n_{MgSO_4}=n_{H_2}=m_{Mg}=0,2\left(mol\right)\)
a, \(m_{MgSO_4}=0,2.120=24\left(g\right)\)
b, \(V_{H_2}=0,2.24,79=4,958\left(l\right)\)
c, Bạn bổ sung đề phần này nhé.
Bài 2 :
\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)
PTHH :
\(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)
0,1 0,3 0,1 0,3
\(m_{Fe_2\left(SO_4\right)_3}=0,1.400=40\left(g\right)\)
\(b,V_{ddH_2SO_4}=\dfrac{0,3}{2}=0,15\left(l\right)\)
\(c,C_{M\left(Fe_2\left(SO_4\right)_3\right)}=\dfrac{0,1}{0,15}=\dfrac{2}{3}\left(M\right)\)
Bài 3 :
\(n_{Mg}=\dfrac{4.8}{24}=0,2\left(mol\right)\)
PTHH :
\(Mg+H_2SO_4\rightarrow MgSO_4+H_2O\)
0,2 0,2 0,2 0,2
\(m_{MgSO_4}=0,2.120=24\left(g\right)\)
\(V_{H_2}=0,2.24,79=4,958\left(l\right)\)
\(c,C_{M\left(H_2SO_4\right)}=\dfrac{0,2}{0,2}=1\left(M\right)\)
\(d,C_{M\left(MgSO_4\right)}=\dfrac{0,2}{0,2}=1\left(M\right)\)
Bài 4 :
\(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\)
PTHH :
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)
0,3 0,45 0,15 0,45
\(V_{H_2}=0,45.24,79=11,1555\left(l\right)\)
\(m_{H_2SO_4}=0,45.98=44,1\left(g\right)\)
\(C\%_{H_2SO_4}=\dfrac{44,1}{300}.100\%=14,7\%\)
\(m_{Al_2\left(SO_4\right)_3}=0,15.342=51,3\left(g\right)\)
\(m_{dd}=8,1+300-\left(0,45.2\right)=307,2\left(g\right)\)
\(C\%_{Al_2\left(SO_4\right)_3}=\dfrac{51,3}{307,2}.100\%\approx16,7\%\)
Bài 5 :
\(n_{H_2}=\dfrac{4,958}{24,79}=0,2\left(mol\right)\)
PTHH:
\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
0,2 0,4 0,2 0,2
\(m_{Fe}=0,2.56=11,2\left(g\right)\)
\(C_{M\left(HCl\right)}=\dfrac{0,4}{0,2}=2\left(M\right)\)
\(C_{M\left(FeCl_2\right)}=\dfrac{0,2}{0,2}=1\left(M\right)\)
nMg=4,824=0,2(mol)nMg=4,824=0,2(mol)
PTHH: Mg + H2SO4 → MgSO4 + H2
Mol: 0,2 0,2
VH2=0,2.24,79=4,958(l)
=> đáp án a
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
PTHH: Mg + H2SO4 → MgSO4 + H2
Mol: 0,2 0,2
\(V_{H_2}=0,2.24,79=4,958\left(l\right)\)
⇒ Chọn C
\(4.a/n_{Al}=\dfrac{5,4}{27}=0,2mol\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
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
\(C_{\%HCl}=\dfrac{0,2.36,5}{200}\cdot100=3,65\%\\ b/C_{\%MgCl_2}=\dfrac{0,1.95}{200+4}\cdot100=4,66\%\\ c/NaOH+HCl\rightarrow NaCl+H_2O\\ n_{NaOH}=n_{HCl}=0,2mol\\ V_{NaOH}=\dfrac{0,2}{1}=0,2l=200ml\)
\(n_{Mg}=\dfrac{7,2}{24}=0,3\left(mol\right)\)
Pt : \(Mg+H_2SO_4\rightarrow MgSO_4+H_2|\)
1 1 1 1
0,3 0,3 0,3
a) \(n_{H2}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,3.22,4=6,72\left(l\right)\)
b) \(n_{H2SO4}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)
\(m_{H2SO4}=0,3.98=29,4\left(g\right)\)
\(m_{ddH2SO4}=\dfrac{29,4.100}{9,8}=300\left(g\right)\)
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