a) \(n_{Fe}=\dfrac{m}{M}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
\(n_{HCl}=C_M.V=3.0,35=1,05\left(mol\right)\)
Lập tỉ lệ: \(n_{Fe}:n_{HCl}=\dfrac{0,2}{1}:\dfrac{1,05}{2}=0,2:0,525\)
=> HCl dư
\(\dfrac{Fe}{0,2}+\dfrac{2HCl}{0,4}->\dfrac{FeCl2}{0,2}+\dfrac{H2}{0,2}\)
\(V_{H_2}=22,4.n=22,4.0,2=4,48\left(l\right)\)
b) \(m_{ddspu}=1,1.350+11,2-0,2.2=395,8\left(g\right)\)
\(m_{FeCl_2}=0,2.127=25,4\left(g\right)\)
\(m_{HCl\left(dư\right)}=\left(1,05-0,4\right).36,5=23,725\left(g\right)\)
\(C\%_{FeCl_2}=\dfrac{25,4}{395,8}.100\approx6,42\%\)
\(C\%_{HCl\left(dư\right)}=\dfrac{23,725}{395,8}.100\approx6\%\)
c) Thừa :v
Có gì không hiểu ib nha bạn ^^
Fe + 2HCl → FeCl2 + H2
\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
\(n_{HCl}=0,35\times3=1,05\left(mol\right)\)
Theo PT: \(n_{Fe}=\dfrac{1}{2}n_{HCl}\)
Theo bài: \(n_{Fe}=\dfrac{4}{21}n_{HCl}\)
Vì \(\dfrac{4}{21}< \dfrac{1}{2}\) ⇒ HCl dư
a) Theo PT: \(n_{H_2}=n_{Fe}=0,2\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,2\times22,4=4,48\left(l\right)\)
b) \(m_{ddHCl}=1,1\times350=385\left(g\right)\)
\(m_{H_2}=0,2\times2=0,4\left(g\right)\)
\(\Rightarrow m_{dd}saupư=11,2+385-0,4=395,8\left(g\right)\)
Dung dịch sau phản ứng gồm: HCl dư và FeCl2
Theo PT: \(n_{HCl}pư=2n_{Fe}=2\times0,2=0,4\left(mol\right)\)
\(\Rightarrow n_{HCl}dư=1,05-0,4=0,65\left(mol\right)\)
\(\Rightarrow m_{HCl}dư=0,65\times36,5=23,725\left(g\right)\)
\(\Rightarrow C\%_{HCl}dư=\dfrac{23,725}{395,8}\times100\%=5,99\%\)
Theo PT: \(n_{FeCl_2}=n_{Fe}=0,2\left(mol\right)\)
\(\Rightarrow m_{FeCl_2}=0,2\times127=25,4\left(g\right)\)
\(\Rightarrow C\%_{FeCl_2}=\dfrac{25,4}{395,8}\times100\%=6,42\%\)
