ta có: nMg = \(\dfrac{12}{24}=0,5\left(mol\right)\)
PTHH: Mg + 2HCl ---> MgCl2 + H2.
Đổi 500ml = 0,5 lít
Ta có: CMHCl = \(\dfrac{n_{HCl}}{0,5}=0,02M\)
=> nHCl = 0,01(mol)
Ta thấy: \(\dfrac{0,5}{1}>\dfrac{0,01}{2}\)
=> Mg dư
Theo PT: \(n_{H_2}=\dfrac{1}{2}.n_{HCl}=\dfrac{1}{2}.0,01=0,005\left(mol\right)\)
=> \(V_{H_2}=0,05.22,4=0,112\left(lít\right)\)