Câu hỏi của Gia Hưng 7P1 STT 10

Question

Lê Ng Hải Anh · Accepted Answer

PT: $Fe+H_2SO_4\rightarrow FeSO_4+H_2$

Ta có: $n_{Fe}=\dfrac{6,72}{56}=0,12\left(mol\right)$

a, $n_{H_2}=n_{Fe}=0,12\left(mol\right)\Rightarrow V_{H_2}=0,12.24,79=2,9748\left(l\right)$

b, $n_{FeSO_4}=n_{Fe}=0,12\left(mol\right)\Rightarrow m_{FeSO_4}=0,12.152=18,24\left(g\right)$

c, PT: $Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O$

$n_{Fe}=\dfrac{2}{3}n_{H_2}=0,08\left(mol\right)\Rightarrow m_{Fe}=0,08.56=4,48\left(g\right)$

Câu trả lời:

PT: $Fe+H_2SO_4\rightarrow FeSO_4+H_2$

Ta có: $n_{Fe}=\dfrac{6,72}{56}=0,12\left(mol\right)$

a, $n_{H_2}=n_{Fe}=0,12\left(mol\right)\Rightarrow V_{H_2}=0,12.24,79=2,9748\left(l\right)$

b, $n_{FeSO_4}=n_{Fe}=0,12\left(mol\right)\Rightarrow m_{FeSO_4}=0,12.152=18,24\left(g\right)$

c, PT: $Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O$

$n_{Fe}=\dfrac{2}{3}n_{H_2}=0,08\left(mol\right)\Rightarrow m_{Fe}=0,08.56=4,48\left(g\right)$