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a, PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
b, \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
Theo PT: \(n_{H_2}=n_{Fe}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)
c, \(n_{H_2SO_4}=n_{Fe}=0,1\left(mol\right)\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,1}{0,2}=0,5\left(M\right)\)
\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\
n_{H_2SO_4}=\dfrac{24,5}{98}=0,25\left(mol\right)\\
pthh:Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)
\(LTL:\dfrac{0,2}{1}< \dfrac{0,25}{1}\)
=> H2SO4 dư
\(n_{H_2}=n_{H_2SO_4\left(p\text{ư}\right)}=n_{Fe}=0,2\left(mol\right)\\
V_{H_2}=0,2.22,4=4,48l\\
m_{H_2SO_4\left(d\right)}=\left(0,25-0,2\right).98=4,9g\)
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ pthh:Mg+2HCl\rightarrow MgCl_2+H_2\)
0,2 0,4 0,2
\(\Rightarrow m_{HCl}=0,4.36,5=14,6\left(g\right)\\ V_{H_2}=0,2.22,4=4,48\left(l\right)\\ pthh:FeO+H_2\underrightarrow{t^o}Fe+H_2O\)
0,2 0,2 0,2
\(m_{Fe}=0,2.56=11,2\left(g\right)\)
a)
\(Mg + H_2SO_4 \to MgSO_4 + H_2\\ n_{H_2} = n_{Mg} = \dfrac{3,6}{24} = 0,15(mol)\\ b)\\ CuO + H_2 \xrightarrow{t^o} Cu + H_2O\\ n_{Cu} = n_{H_2} = 0,15(mol)\\ \Rightarrow m_{Cu} = 0,15.64 = 9,6(gam)\)
\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\a, PTHH:Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ b,n_{H_2}=n_{H_2SO_4}=n_{Fe}=0,2\left(mol\right)\\ V_{H_2\left(đkc\right)}=0,2.24,79=4,958\left(l\right)\\ c,C_{MddH_2SO_4}=\dfrac{0,2}{0,4}=0,5\left(M\right)\)
\(n_{Mg}=\dfrac{10,8}{24}=0,45\left(mol\right)\\
pthh:Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\)
0,45 0,45 0,45
\(m_{H_2SO_4}=0,45.98=44,1\left(g\right)\\
C\%_{H_2SO_4}=\dfrac{44,1}{176,4}.100\%=25\%\\
V_{H_2}=0,45.22,4=10,08\left(l\right)\)
a, \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
b, \(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\)
Theo PT: \(n_{H_2}=n_{Zn}=0,3\left(mol\right)\Rightarrow V_{H_2}=0,3.22,4=6,72\left(l\right)\)
c, \(n_{ZnSO_4}=n_{Zn}=0,3\left(mol\right)\Rightarrow m_{ZnSO_4}=0,3.161=48,3\left(g\right)\)
d, \(n_{Fe_2O_3}=\dfrac{64}{160}=0,4\left(mol\right)\)
PT: \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
Xét tỉ lệ: \(\dfrac{0,4}{1}>\dfrac{0,3}{3}\), ta được Fe2O3 dư.
Mà: H% = 30% \(\Rightarrow n_{H_2\left(pư\right)}=0,3.30\%=0,09\left(mol\right)\)
Theo PT: \(\left\{{}\begin{matrix}n_{Fe}=\dfrac{2}{3}n_{H_2}=0,06\left(mol\right)\\n_{Fe_2O_3\left(pư\right)}=\dfrac{1}{3}n_{H_2}=0,03\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow n_{Fe_2O_3\left(dư\right)}=0,4-0,03=0,37\left(mol\right)\)
\(\Rightarrow a=m_{Fe}+m_{Fe_2O_3\left(dư\right)}=62,56\left(g\right)\)
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ a,Mg+2HCl\rightarrow MgCl_2+H_2\\ n_{MgCl_2}=n_{H_2}=n_{Mg}=0,2\left(mol\right);n_{HCl}=0,2.2=0,4\left(mol\right)\\ b,C_{MddHCl}=\dfrac{0,4}{0,1}=4\left(M\right)\\ V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\)
1)
4P + 5O2 --to--> 2P2O5
Fe2O3 + 3H2 --to-> 2Fe + 3H2O
2KClO3 ---to--> 2KCl + 3O2
2) a) Phương trình hóa học phản ứng
Mg + H2SO4 ----> MgSO4 + H2
b) \(n_{Mg}=\frac{m}{M}=\frac{9,6}{24}=0,4\left(mol\right)\)
\(m_{H_2SO_4}=D.V=250.1,83=457,5\left(g\right)\)
\(n_{H_2SO_4}=\frac{m}{M}=\frac{457,5}{98}=4.7\left(mol\right)\)
Nhận thấy \(\frac{0,4}{1}< \frac{4,7}{1}\)
=> H2SO4 dư
=> \(V_{H_2}=n.22,4=0,4.22,4=8,96\left(l\right)\)
c) \(C_{mol}=\frac{n}{V}=\frac{4,7}{0,25}=18,8M\)