Question

Bài 32 giải giúp tớ với

Answer 1

a) PTHH:

\(Fe+2HCl->FeCl_2+H_2\)

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

b) Dựa vào PTHH:

Ta thấy \(n_{H_2}=n_{Fe}=0,2\left(mol\right)\)

-> \(V_{H_2\left(đktc\right)}=n_{H_2}\cdot22,4=0,2\cdot22,4=4,48\left(l\right)\)

Muối thu được là \(FeCl_2\)

ta thấy \(n_{Fe}=n_{FeCl_2}=0,2\left(mol\right)\)

--> \(m_{FeCl_2}=n_{FeCl_2}\cdot M_{FeCl_2}=0,2\cdot127=25,4\left(g\right)\)

c) Ta thấy \(n_{HCl}=2\cdot n_{Fe}=2\cdot0,2=0,4\left(mol\right)\)

-> \(m_{HCl}=n_{HCl}\cdot M_{HCl}=0,4\cdot36,5=14,6\left(g\right)\)

d) \(C\%_{HCl}=\dfrac{m_{HCl}}{m_{ddHCl}}\cdot100\%=\dfrac{14,6}{200}\cdot100\%=7,3\%\)