Question

Cho 2,8g sắt tác dụng với 250g H2SO4 thu dc dd muối có nòng độ 2M.Hãy tính;

C% dd axit tham gia

Thể tích dd muối thu dc

Thể tích khí hidro sinh ra ở dktc

Answer 1

\(n_{Fe}=\dfrac{2,8}{56}=0,05\left(mol\right)\)

PTHH: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

          0,05--->0,05----->0,05--->0,05

\(m_{H_2SO_4}=0,05.98=4,9\left(g\right)\Rightarrow C\%_{H_2SO_4}=\dfrac{4,9}{250}.100\%=1,96\%\)

\(V_{dd.sau.pư}=\dfrac{0,05}{2}=0,025\left(l\right)\)

\(V_{H_2}=0,05.22,4=1,12\left(l\right)\)

Answer 2

\(n_{Fe}=\dfrac{2,8}{56}=0,05\left(mol\right)\)

PTHH: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

         0,05--->0,05------->0,05----->0,05

=> \(\left\{{}\begin{matrix}C\%_{H_2SO_4}=\dfrac{0,05.98}{250}.100\%=1,96\%\\V_{dd.FeSO_4}=\dfrac{0,05}{2}=0,025\left(l\right)=25\left(ml\right)\\V_{H_2}=0,05.22,4=1,12\left(l\right)\end{matrix}\right.\)

Answer 3

Ta có: \(n_{Fe}=\dfrac{2,8}{56}=0,05\left(mol\right)\)

PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

___0,05___0,05_____0,05___0,05 (mol)

\(m_{H_2SO_4}=0,05.98=4,9\left(g\right)\Rightarrow C\%_{H_2SO_4}=\dfrac{4,9}{250}.100\%=1,96\%\)

\(V_{ddFeSO_4}=\dfrac{0,05}{2}=0,025\left(l\right)\)

\(V_{H_2}=0,05.22,4=1,12\left(l\right)\)

Bạn tham khảo nhé!