a) \(3Fe+2O_2\underrightarrow{t^0}Fe_3O_4\left(1\right)\)
b) \(n_{Fe}=\dfrac{126}{56}=2,25\left(mol\right)\)
PTHH: \(3Fe+2O_2\underrightarrow{t^0}Fe_3O_4\left(1\right)\)
Theo PTHH: \(n_{Fe}:n_{O_2}=3:2\)
\(\Rightarrow n_{O_2}=n_{Fe}.\dfrac{2}{3}=2,25.\dfrac{2}{3}=1,5\left(mol\right)\)
\(\Rightarrow V_{O_2}=1,5.22,4=33,6\left(l\right)\)
c) PTHH: \(2KClO_3\underrightarrow{t^0}2KCl+3O_2\uparrow\)
Ta có: \(n_{O_2}=1,5\left(mol\right)\) ( câu b)
Theo PTHH: \(n_{O_2}:n_{KClO_3}=3:2\)
\(\Rightarrow n_{KClO_3}=n_{O_2}.\dfrac{2}{3}=1,5.\dfrac{2}{3}=1\left(mol\right)\)
\(\Rightarrow m_{KClO_3}=1.122,5=122,5\left(g\right)\)
a. PTHH: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\\ 2,25mol:1,5mol\rightarrow0,75mol\)
\(n_{Fe}=\dfrac{126}{56}=2,25\left(mol\right)\)
b. \(V_{O_2}=1,5.22,4=33,6\left(l\right)\)
c. Sửa: Kali Clorat nhé bạn!
PTHH: \(2KClO_3\xrightarrow[MnO_2]{t^o}2KCl+3O_2\\ 1mol\leftarrow1mol:1,5mol\)
\(m_{KClO_3}=122,5.1=122,5\left(g\right)\)
3Fe + 2O2(1) -------> Fe3O4
2,25 1,(6)
2KClO3 ------->2KCl + 3O2(2)
1,(6)
nFe= 126/56 = 2,25 mol
=> nO2 (1)= 2,25 x 2/3 =1,(6) mol
Vì VO2 (1) = VO2 (2) => số mol của chúng sẽ bằng nhau
=> nKClO3 = 1,(6) x 2/3 =1.(1) mol
=> mKClO3= 1,(1) x (39+ 35,5+ 16x3)= 136,(1) gam
\(n_{Fe}=\dfrac{126}{56}=2,25\left(mol\right)\)
a. PTHH: \(3Fe+2O_2-t^o->Fe_3O_4\left(1\right)\)
b. Theo PT (1) ta có: \(n_{O_2}=\dfrac{2,25.2}{3}=1,5\left(mol\right)\)
\(\Rightarrow V_{O_2}=1,5.22,4=33,6\left(l\right)\)
c. PTHH: \(2KClO_3-t^o->2KCl+3O_2\uparrow\left(2\right)\)
Theo PT (2) ta có: \(n_{KCl}=\dfrac{1,5.2}{3}=1\left(mol\right)\)
\(\Rightarrow m_{KCl\left(cầndùng\right)}=1.74,5=74,5\left(g\right)\)
