Question

Giải giúp mình với ạ

Answer 1

Fe+S-to>FeS

 0,2---------0,2 mol

n _Fe=\(\dfrac{11,2}{56}\)=0,2 mol

n _S=\(\dfrac{9,6}{32}\)=0,3 mol

=>S dư 

=>m _S=0,1.32=3,2g

=>m _FeS=0,2.88=17,6g

Answer 2

Answer 3

a) PTHH : Fe + S ->  FeS

b) \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ n_S=\dfrac{9,6}{32}=0,3\left(mol\right)\)

\(\Rightarrow\dfrac{n_{Fe}}{1}< \dfrac{n_S}{1}\) => S dư

=> \(m_S=0,1.32=3,2\left(g\right)\)

c) \(m_{FeS}=0,2.88=17,6\left(g\right)\)