Question

Cho 1 khối lượng mạt sắt dư tương đương vs 200g dung dịch H2SO4 20% a. vt pthh b. Tính khối lượng sắt phản ứng c. Tính thể tích khi sinh ra (đktc)

Answer 1

\(n_{H_2SO_4}=\dfrac{200.20\%}{98}=\dfrac{20}{49}\left(mol\right)\\a, PTHH:Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ n_{Fe}=n_{H_2}=n_{H_2SO_4}=\dfrac{20}{49}\left(mol\right)\\ b,m_{Fe}=\dfrac{20}{49}.56=\dfrac{160}{7}\left(g\right)\\ c,V_{H_2\left(đktc\right)}=\dfrac{20}{49}.22,4=\dfrac{64}{7}\left(l\right)\)

Answer 2

a, PT: \(Fe+2H_2SO_4\rightarrow FeSO_4+H_2\)

b, \(m_{H_2SO_4}=200.20\%=40\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{40}{98}=\dfrac{20}{49}\left(mol\right)\)

Theo PT: \(n_{Fe}=n_{H_2SO_4}=\dfrac{20}{49}\left(mol\right)\Rightarrow m_{Fe}=\dfrac{20}{49}.56=\dfrac{160}{7}\left(g\right)\)

c, \(n_{H_2}=n_{H_2SO_4}=\dfrac{20}{49}\left(mol\right)\Rightarrow V_{H_2}=\dfrac{20}{49}.22,4=\dfrac{64}{7}\left(l\right)\)