a, \(\dfrac{3}{2}\)nAl = a +\(\dfrac{3}{2}\)\(\dfrac{2,4}{0,1}\) \(\dfrac{16}{80}\)=0,2(mol)
PTHH: CuO + H2 ---to---> Cu + H2O
LTL: 0,2 < 0,85 => H2 dư
Gọi nCuO (pư) = a (mol)
=> nCu (sinh ra) = a (mol)
Ta có: mchất rắn (sau pư) = 80(0,2 - a) + 64a = 13,6
=> a = 0,15 (mol)
=> H =
