a) \(n_{Al}=\dfrac{m}{M}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
\(PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
Số mol : 2mol.........6mol
Sau p/ư : 0,2mol........ x
\(n_{HCl}=x=\dfrac{0,2.6}{2}=0,6\left(mol\right)\)
Ta có : \(n_{HCl}=\dfrac{16\%.m_{HCl}}{100\%.36,5}\)
\(\Rightarrow0,6=\dfrac{16.m_{HCl}}{100.36,5}\)
\(\Rightarrow m_{HCl}=\dfrac{2190}{16}=136,875\left(g\right)\)
b) \(n_{AlCl_3}=\dfrac{3}{2}n_{Al}=0,3\left(mol\right)\)
\(m_{AlCl_3}=0,3.133,5=40,05\left(g\right)\)
\(\Rightarrow C_{\%ddAlCl_3}=\dfrac{40,05}{5,4+136,875}.100\approx27,95\%\)
nAl=5,4/27=0,2(mol)
PT:
2Al + 6HCl -> 2AlCl3 + 3H2↑
0.2........0.6........0.2..........0.3 (mol)
a) mHCl=0,6.36,5=21,9 (g)
b)mAlCl3=0.2*133.5=26.7(g)
=> C% (AlCl3)=(26.7/(21.9*100/16+5.4))*100%=18.766%
nAl = 0,2 mol
2Al + 6HCl \(\rightarrow\) 2AlCl3 + 3H2
0,2.....0,6...........0,6..........0,3
\(\Rightarrow\) mHCl đã dùng = \(\dfrac{0,6.36,5.100}{16}\) = 136,875 (g)
\(\Rightarrow\) mdd sau phản ứng = 5,4 + 136,875 - ( 0,3.2 ) = 141,675 (g)
\(\Rightarrow\) C%dd muối sau phản ứng = \(\dfrac{0,2.133,5.100}{141,675}\) \(\approx\) 18,8%
