Bài 6:
a. $2Al_{ }$ + $6HCl_{ }$ → $2AlCl_{3}$ + $3H_{2}↑$
b. $n_{H{2}}$ = 6.72/22.4 = 0.3 mol
⇒ $n_{Al}$ = $2/3n_{H{2}}$ = 2/3*0.3 = 0.2 mol
⇒ $a_{ }$ = $m_{Al}$ = 0.2*27 =5.4 g
c. Ta có: $n_{HCl}$ = $6/3n_{H{2}}$ = 0.6 mol
⇒ $V_{HCl}$ = 0.6/2 = 0.3 lít
Câu 7:
a. $Fe_{ }$ + $2HCl_{ }$ → $FeCl_{2}$ + $H_{2}↑$
b. $n_{HCl}$ = [(200*14.6)/100]/36.5 = 0.8 mol
⇒ $n_{Fe}$ = $1/2n_{HCl}$ = 1/2*0.8 = 0.4 mol
⇒ $a_{ }$ = $m_{Fe}$ = 0.4*56 = 22.4 g
c. $n_{H{2}}$ = $1/2n_{HCl}$ = 1/2*0.8 = 0.4 mol
⇒ $V_{H{2}}$ = 0.4*22.4 =8.96 lít
