\(n_{Fe}=\dfrac{2,8}{56}=0,05\left(mol\right)\)
Pt : \(Fe+2HCl\rightarrow FeCl_2+H_2|\)
1 2 1 1
0,05 0,1
\(n_{HCl}=\dfrac{0,05.2}{1}=0,1\left(mol\right)\)
\(m_{HCl}=0,1.36,5=3,65\left(g\right)\)
\(m_{ddHCl}=\dfrac{3,65.100}{3,65}=100\left(g\right)\)
Chúc bạn học tốt
\(n_{Fe}=\dfrac{2,8}{56}=0,05\left(mol\right)\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ \Rightarrow n_{HCl}=2n_{Fe}=0,1\left(mol\right)\Rightarrow m_{HCl}=0,1\cdot36,5=3,65\left(g\right)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{3,65\cdot100\%}{3,65\%}=100\left(g\right)\)
Ta có: \(n_{Fe}=\dfrac{2,8}{56}=0,05\left(mol\right)\)
\(PTHH:Fe+2HCl--->FeCl_2+H_2\uparrow\)
Theo PT: \(n_{HCl}=2.n_{Fe}=2.0,05=0,1\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,1.36,5=3,65\left(g\right)\)
\(\Rightarrow m_{dd_{HCl}}=\dfrac{3,65.100\%}{3,65\%}=100\left(g\right)\)
