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\(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)
PTHH :
\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\)
0,1 0,1 0,1 0,1
\(a,m_{MgSO_4}=0,1.120=12\left(g\right)\)
\(b,V_{H_2}=0,1.22,4=2,24\left(l\right)\)
\(c,m_{ddH_2SO_4}=\dfrac{0,1.98.100}{10}=98\left(g\right)\)
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
PTHH: Mg + 2HCl → MgCl2 + H2
Mol: 0,2 0,4 0,2
\(V_{H_2}=0,2.24,79=4,958\left(l\right)\)
\(m_{HCl}=0,4.36,5=14,6\left(g\right)\Rightarrow m_{ddHCl}=\dfrac{14,6.100\%}{20\%}=73\left(g\right)\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\\ n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ n_{H_2}=n_{Mg}=0,2\left(mol\right)\\ V_{H_2\left(25^oC,1bar\right)}=0,2.22,4=4,48\left(l\right)\\ n_{HCl}=0,2.2=0,4\left(mol\right)\\ m_{HCl}=0,4.36,5=14,6\left(g\right)\)
a, \(n_{Fe}=\dfrac{0,56}{56}=0,01\left(mol\right)\)
PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
Theo PT: \(n_{FeSO_4}=n_{H_2}=n_{Fe}=0,01\left(mol\right)\)
\(\Rightarrow m_{FeSO_4}=0,01.152=1,52\left(g\right)\)
\(V_{H_2}=0,01.22,4=0,224\left(l\right)\)
b, \(n_{H_2SO_4}=n_{Fe}=0,01\left(mol\right)\Rightarrow m_{ddH_2SO_4}=\dfrac{0,01.98}{19,6\%}=5\left(g\right)\)
c, Ta có: m dd sau pư = 0,56 + 5 - 0,01.2 = 5,54 (g)
\(\Rightarrow C\%_{FeSO_4}=\dfrac{1,52}{5,54}.100\%\approx27,44\%\)
\(a,n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ b,n_{MgCl_2}=n_{H_2}=n_{Mg}=0,2\left(mol\right)\\ b,V_{H_2\left(25\text{đ}\text{ộ}C,1bar\right)}=0,2.24,79=4,958\left(l\right)\\ c,n_{HCl}=2.0,2=0,4\left(mol\right)\\ m_{\text{dd}HCl}=\dfrac{0,4.36,5.100}{10}=146\left(g\right)\\ m_{\text{dd}A}=m_{Mg}+m_{\text{dd}HCl}-m_{H_2}=4,8+146-0,2.2=150,4\left(g\right)\\ d,C\%_{\text{dd}MgCl_2}=\dfrac{95.0,2}{150,4}.100\approx12,633\%\)
\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\
pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,1 0,2 0,1 0,1
\(m_{HCl}=0,2.36,5=7,3\left(g\right)\\
V_{H_2}=0,1.22,4=2,24l\\
m_{\text{dd}}=6,5+200-\left(0,1.2\right)=206,3g\)
bài 2 :
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\
pthh:Mg+2HCl\rightarrow MgCl_2+H_2\)
0,2 0,4 0,2 0,2
\(m_{HCl}=0,4.36,5=14,6g\\
V_{H_2}=0,2.22,4=4,48l\\
m\text{dd}=4,8+200-0,4=204,4g\\
C\%=\dfrac{0,2.136}{204,4}.100\%=13,3\%\)
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ Mg+H_2SO_4\rightarrow MgSO_4+H_2\\ n_{MgSO_4}=n_{H_2}=n_{H_2SO_4}=n_{Mg}=0,2\left(mol\right)\\ a,m_{MgSO_4}=120.0,2=24\left(g\right)\\ b,V_{H_2\left(đkc\right)}=24,79.0,2=4,958\left(l\right)\\ c,Oxide:A_2O_x\left(x:hoá.trị.A\right)\\ A_2O_x+xH_2SO_4\rightarrow A_2\left(SO_4\right)_x+xH_2O\\ n_{Oxide}=\dfrac{\dfrac{3}{4}.0,2.1}{x}=\dfrac{0,15}{x}\left(mol\right)\\ M_{A_2O_x}=\dfrac{8}{\dfrac{0,15}{x}}=\dfrac{160}{3}x\)
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
\(n_{Mg}=\dfrac{7,2}{24}=0,3\left(mol\right)\)
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) \(n_{H2SO4}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)
\(m_{H2SO4}=0,3.98=29,4\left(g\right)\)
\(m_{ddH2SO4}=\dfrac{29,4.100}{9,8}=300\left(g\right)\)
Chúc bạn học tốt
a)
\(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\)
PTHH: Fe + 2HCl --> FeCl2 + H2
0,15-->0,3----->0,15-->0,15
=> \(V_{H_2}=0,15.22,4=3,36\left(l\right)\)
b) \(m_{FeCl_2}=0,15.127=19,05\left(g\right)\)
c) \(m_{HCl}=0,3.36,5=10,95\left(g\right)\)
=> \(C\%=\dfrac{10,95}{200}.100\%=5,475\%\)
