Zn + 2HCl -> ZnCl2 + H2 (1)
nZn=0,1(mol)
nHCl=0,1(mol)
Vì \(\dfrac{0,1}{2}< 0,1\) nên sau PƯ Zn dư
Theo PTHH 1 ta có:
nZnCl2=\(\dfrac{1}{2}\)nHCl=0,05(mol)
mZnCl2=136.0,05=6,8(g)
c;
2HCl + Ca(OH)2 -> CaCl2 + 2H2O (2)
nHCl=0,2(mol)
Từ 2:
nCa(OH)2=\(\dfrac{1}{2}\)nHCl=0,1(mol)
VCa(OH)2=\(\dfrac{0,1}{2}=0,05\left(lít\right)\)