a) 2Al + 6HCl --> 2AlCl3 + 3H2
b) \(n_{HCl}=\dfrac{100.7,3}{100.36,5}=0,2\left(mol\right)\)
PTHH: 2Al + 6HCl --> 2AlCl3 + 3H2
____\(\dfrac{1}{15}\)<---0,2------->\(\dfrac{1}{15}\)
=> \(m_{Al}=\dfrac{1}{15}.27=1,8\left(g\right)\)
c) \(m_{AlCl_3}=\dfrac{1}{15}.133,5=8,9\left(g\right)\)
2Al + 6HCl -> 2AlCl3 + 3H2
1/15 0.2 1/15
mct HCl = 7.3% × 100 = 7.3 g => nHCl = 0.2mol
mAl = 1/15 × 27 = 1.8g
mAlCl3 = 1 15 × 133.5 = 8.9 g