Question

Đốt nóng hoàn toàn 11,2g Fe trong bình chứa oxi tạo thành oxit sắt từ

a. tính V khí oxi đo ở ĐKTC đã t/gia PỨ

b. tính KL oxit sắt từ

Answer 1

\(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)

a) \(n_{Fe}=\dfrac{m}{M}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

Theo PTHH, \(n_{O_2}=\dfrac{2}{3}n_{Fe}=\dfrac{2}{3}\cdot0,2=\dfrac{2}{15}\left(mol\right)\)

\(V_{O_2}=n\cdot22,4=\dfrac{2}{15}\cdot22,4=\dfrac{224}{75}\left(l\right)\)

b) Theo PTHH, \(n_{Fe_3O_4}=\dfrac{4}{3}n_{Fe}=\dfrac{4}{3}\cdot0,2=\dfrac{4}{15}\left(mol\right)\)

\(m_{Fe_3O_4}=n\cdot M=\dfrac{4}{15}\cdot232=\dfrac{928}{15}\left(g\right)\)

Answer 2

$3Fe+2O_2\underrightarrow{t^o}F{e}_3{O}_4$

a) $n_{Fe}=\frac{m}{M}=\frac{11,2}{56}=0,2\left(mol\right)$

Theo PTHH, $n_{O_2}=\frac{2}{3}{n}_{Fe}=\frac{2}{3}\cdot 0,2=\frac{2}{15}\left(mol\right)$

$V_{O_2}=n\cdot 22,4=\frac{2}{15}\cdot 22,4=\frac{224}{75}\left(l\right)$

b) Theo PTHH, $n_{F{e}_3{O}_4}=\frac{4}{3}{n}_{Fe}=\frac{4}{3}\cdot 0,2=\frac{4}{15}\left(mol\right)$

$m_{F{e}_3{O}_4}=n\cdot M=\frac{4}{15}\cdot 232=\frac{928}{15}\left(g\right)$

Answer 3

a) \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

PTHH: \(3Fe+2O_2\underrightarrow{t^0}Fe_3O_4\)

Theo PTHH: \(n_{Fe}:n_{O_2}=3:2\)

\(\Rightarrow n_{O_2}=n_{Fe}.\dfrac{2}{3}=0,2.\dfrac{2}{3}=\dfrac{2}{15}\left(mol\right)\)

\(\Rightarrow V_{O_2\left(đktc\right)}=\dfrac{2}{15}.22,4=3\left(l\right)\)

b) Theo PTHH: \(n_{Fe}:n_{Fe_3O_4}=3:1\)

\(\Rightarrow n_{Fe_3O_4}=n_{Fe}.\dfrac{1}{3}=0,2.\dfrac{1}{3}=\dfrac{1}{15}\left(mol\right)\)

\(\Rightarrow m_{Fe_3O_4}=\dfrac{1}{15}.232=15,4\left(g\right)\)