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\(n_{O_2}=\dfrac{3,7185}{24,79}=0,15\left(mol\right)\\ 3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\\ n_{Fe}=\dfrac{3}{2}.n_{O_2}=1,5.0,15=0,225\left(mol\right)\\ \Rightarrow m_{Fe}=0,225.56=12,6\left(g\right)\\ n_{Fe_3O_4}=\dfrac{n_{O_2}}{2}=\dfrac{0,15}{2}=0,075\left(mol\right)\\ \Rightarrow m_{Fe_3O_4}=232.0,075=17,4\left(g\right)\)
nFe2O3 = 32/160 = 0,2 (mol)
PTHH: Fe2O3 + 3H2 -> (t°) 2Fe + 3H2O
Mol: 0,2 ---> 0,6 ---> 0,4
mFe = 0,4 . 56 = 22,4 (g)
VH2 = 0,6 . 24,79 = 14,874 (l)
\(a,n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\\ 3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\\ b,n_{Fe_3O_4}=\dfrac{n_{Fe}}{3}=\dfrac{0,3}{3}=0,1\left(mol\right)\\ \Rightarrow m_{Fe_3O_4}=232.0,1=23,2\left(g\right)\\ n_{O_2}=\dfrac{2}{3}.n_{Fe}=\dfrac{2}{3}.0,3=0,2\left(mol\right)\\ \Rightarrow V_{O_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\)
a)
\(n_{H_2}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\)
PTHH: 2Al + 6HCl --> 2AlCl3 + 3H2
0,2<---0,6<-----0,2<---0,3
=> mAl = 0,2.27 = 5,4 (g)
mHCl = 0,6.36,5 = 21,9 (g)
b) mAlCl3 = 0,2.133,5 = 26,7 (g)
c)
PTHH: Fe2O3 + 3H2 --to--> 2Fe + 3H2O
0,1<---0,3---------->0,2
=> mFe2O3 = 0,1.160 = 16 (g)
d) mFe = 0,2.56 = 11,2 (g)
a.b.\(n_{H_2}=\dfrac{V_{H_2}}{24,79}=\dfrac{7,437}{24,79}=0,3mol\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,2 0,6 0,2 0,3 ( mol )
\(m_{Al}=n_{Al}.M_{Al}=0,2.27=5,4g\)
\(m_{HCl}=n_{HCl}.M_{HCl}=0,6.36,5=21,9g\)
\(m_{AlCl_3}=n_{AlCl_3}.M_{AlCl_3}=0,2.133,5=26,7g\)
c.d.\(Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\)
0,1 0,3 0,2 ( mol )
\(m_{Fe_2O_3}=n_{Fe_2O_3}.M_{Fe_2O_3}=0,1.160=16g\)
\(m_{Fe}=n_{Fe}.M_{Fe}=0,2.56=11,2g\)
Câu 6:
1. \(n_{O_2}=\dfrac{2,470}{24,79}=0,1\left(mol\right)\)
PTHH:
\(4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\)
2/15 0,1 1/15
\(m_{Al}=\dfrac{2}{15}.27=3,6\left(g\right)\)
2. Gọi m KCl cần thêm là x
Ta có:
\(15\%=\dfrac{\dfrac{10x.10}{100}+\dfrac{300.25}{100}}{10x+300}\)
\(\Rightarrow x=60\)
Vậy \(m_{ddKCl}=\dfrac{60.100}{10}=600\left(g\right)\)
a,
\(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
b, \(nFe=\dfrac{16,8}{56}=0,3\left(mol\right)\)
\(\Rightarrow nO_2=0,3.\dfrac{2}{3}=0,2\left(mol\right)\)
\(VO_2=0,2.24,79=4,958\left(l\right)\)
c, \(nFe_3O_4=0,1\left(mol\right)\)
\(mFe_3O_4=0,1.232=23,2\left(gam\right)\)
\(n_{Fe_2O_3}=\dfrac{24}{160}=0,15(mol)\\ \Rightarrow n_{CO}=3n_{Fe_2O_3}=0,45(mol)\\ \Rightarrow V_{CO(25^oC,1bar)}=24,79.0,45=11,1555(l)\)
Chọn B
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