Sửa đề : 500g dd HCl 10%
a)2Al + 6HCl \(\rightarrow\) 2AlCl3 + 3H2
b) mHCl = \(\dfrac{C\%.m_{dd}}{100\%}=\dfrac{10\%.500}{100\%}=50\left(g\right)\)
=> nHCl = 50/36,5 = 100/73 (mol)
Theo PT => nH2 =1/2 . nHCl = 1/2 . 100/73 =50/73 (mol)
=> VH2 = n . 22,4 = 50/73 . 22,4 =15,34(l)
c) Theo PT => nAlCl3 = 1/3 . nHCl = 1/3 . 100/73 =100/219 (mol)
=> mAlCl3 = n .M = 100/219 .133,5 =60,96(g)