4Fe+3O2-to>2Fe2O3

0,8-----0,6------0,4 mol

n Fe=\(\dfrac{44,8}{56}\)=0,8 mol

=>VO2=0,6.22,4=13,44l

=>m Fe2O3=0,4.160=64g

d) 2KMnO4-to>K2MnO4+MnO2+O2

        1,2---------------------------------0,6

=>m KMnO4=1,2.158=189,6g

3Fe+2O2-to>Fe3O4

0,06---0,04------0,02 mol

n Fe3O4=\(\dfrac{4,64}{232}\)=0,02 mol

=>m Fe=0,06.56=3,36g

=>VO2=0,04.22,4=0,896l

b)2KClO3-to>2KCl+3O2

    \(\dfrac{2}{75}\)------------------0,04 mol

=>m KClO3=\(\dfrac{2}{75}\).122,5=32,67g

1.\(n_{Fe_3O_4}=\dfrac{2,32}{232}=0,01mol\)

\(3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\)

0,3                             0,1    ( mol )

\(m_{Fe}=0,3.56=16,8g\)

2.\(n_{Cu}=\dfrac{3,2}{64}=0,05mol\)

\(2Cu+O_2\rightarrow\left(t^o\right)2CuO\)

0,05                       0,05     ( mol )

\(m_{CuO}=0,05.80=4g\)

3.\(n_{Na}=\dfrac{4,6}{23}=0,2mol\)

\(4Na+O_2\rightarrow\left(t^o\right)2Na_2O\)

0,2      0,05                         ( mol )

\(V_{O_2}=0,05.24,79=1,2395l\)

4.\(n_{Cu}=\dfrac{1,6}{64}=0,025mol\)

\(2Cu+O_2\rightarrow\left(t^o\right)2CuO\)

0,025 0,0125                     ( mol )

\(V_{O_2}=0,0125.24,79=0,309875l\)

Giúp mình với

\(a,n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\\ 3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\\ b,n_{Fe_3O_4}=\dfrac{n_{Fe}}{3}=\dfrac{0,3}{3}=0,1\left(mol\right)\\ \Rightarrow m_{Fe_3O_4}=232.0,1=23,2\left(g\right)\\ n_{O_2}=\dfrac{2}{3}.n_{Fe}=\dfrac{2}{3}.0,3=0,2\left(mol\right)\\ \Rightarrow V_{O_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\)

.iron là sắt mà s cho S nx v?

PTPU: Fe + O₂ -> Fe₂O₃

Áp dụng định luật BTKL:

m_Fe + m_O2 = m_Fe2O3

=> mFe2O3 = 23,2 - 16,8 = 6,4g

=> Chọn C

PTHH: \(3Fe+2O_2\overset{t^o}{--->}Fe_3O_4\)

Ta có: \(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)

\(n_{Fe_3O_4}=\dfrac{23,2}{232}=0,1\left(mol\right)\)

Ta thấy: \(\dfrac{0,3}{3}=\dfrac{0,1}{1}\)

=> Không có chất dư.

Theo PT: \(n_{O_2}=2.n_{Fe_3O_4}=2.0,1=0,2\left(mol\right)\)

=> \(m_{O_2}=0,2.32=6,4\left(g\right)\)

Chọn C

3Fe+2O₂-to>Fe₃O₄

0,225---0,15---0,075

n _O2= \(\dfrac{3,7185}{24,79}\)=0,15 mol

=>m _Fe=0,225.56=12,6g

=>m _Fe3O4=0,075.232=17,4g

\(n_{O_2}=\dfrac{3,7185}{24,79}=0,15\left(mol\right)\\ 3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\\ n_{Fe}=\dfrac{3}{2}.n_{O_2}=1,5.0,15=0,225\left(mol\right)\\ \Rightarrow m_{Fe}=0,225.56=12,6\left(g\right)\\ n_{Fe_3O_4}=\dfrac{n_{O_2}}{2}=\dfrac{0,15}{2}=0,075\left(mol\right)\\ \Rightarrow m_{Fe_3O_4}=232.0,075=17,4\left(g\right)\)

đkc đúng k bn :) ?

a) PTHH : 3Fe + 2O₂ \(\underrightarrow{t^0}\) Fe₃O₄

b) Theo ĐLBTKL

\(m_{Fe}+m_{O_2}=m_{Fe_3O_4}\\ =>m_{O_2}=11,6-8,4=3,2\left(g\right)\)