Fe + 2HCl → \(FeCl_2+H_2\)
x...........2x..............x.............x (mol)
2Al + 6HCl → \(2AlCl_3+3H_2\)
y...........3y...............y..............\(\dfrac{3y}{2}\) (mol)
a) \(n_{H_2}=\dfrac{V}{22,4}=\dfrac{5,6}{22,4}\)= 0,25 (mol)
Gọi x (mol) là số mol của Fe
y (mol) là số mol của Al
Ta có : \(\left\{{}\begin{matrix}56x+27y=8,3\\x+\dfrac{3}{2}y=0,25\end{matrix}\right.\)
⇔ \(\left\{{}\begin{matrix}x=0,1\left(mol\right)\\y=0,1\left(mol\right)\end{matrix}\right.\)
\(m_{Fe}\)= n . M = 0,1 . 56 = 5,6 (g)
\(m_{Al}\)= n . M = 0,1 . 27 = 2,7 (g)
\(\%m_{Fe}=\dfrac{m_{Fe}.100}{m_{hh}}=\dfrac{5,6.100}{8,3}\)≃ 67, 47 (%)
\(\%m_{Al}=\dfrac{m_{Al}.100}{m_{hh}}=\dfrac{2,7.100}{8,3}\)≃ 32,53 (%)
b) \(n_{HClbanđầu}=\dfrac{n_{p.ứ}.\left(20+100\right)}{100}\)= \(\dfrac{\left(2.0,1+3.0,1\right).120}{100}\)= 0,6 (mol)
\(n_{HCldư}\)= \(n_{b.đ}-n_{p.ứ}\) = 0,6 - (2 . 0,1 + 3 . 0,1) = 0,1 (mol)
\(m_{HCl}\)= n . M = (2 . 0,1 + 3 . 0,1) . 36,5 = 18,25 (g)
\(m_{ddHCl}=\dfrac{m_{HCl}.100}{C\%}=\dfrac{18,25.100}{7,3}\)= 250 (g)
\(m_{ddsauphảnứng}=m_{hh}+m_{ddHCl}-m_{H_2}\)
= 8,3 + 250 - 0,25 . 2 = 257,8(g)
\(C\%_{FeCl_2}=\dfrac{m_{FeCl_2}.100}{m_{dd}}\)= \(\dfrac{0,1.127.100}{257,8}\) ≃ 4,93 (%)
\(C\%_{AlCl_3}=\dfrac{m_{AlCl_3}.100}{m_{dd}}=\dfrac{0,1.133,5.100}{257,8}\) ≃ 5,18 (%)
\(C\%_{HCldư}=\dfrac{m_{HCl.dư}.100}{m_{ddHCl}}=\dfrac{0,1.36,5.100}{250}\)= 1,46 (%)