PTHH: Mg + H2SO4 (loãng) -> MgSO4 + H2 (1)
Fe + H2SO4 (loãng) -> FeSO4 + H2 (2)
Ta có: \(\left[{}\begin{matrix}n_{Mg\left(1\right)}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\n_{Fe\left(2\right)}=\dfrac{11,2}{56}=0,2\left(mol\right)\end{matrix}\right.\)
Theo PTHH và đề bài, ta có:
\(\left[{}\begin{matrix}n_{H_2\left(1\right)}=n_{Mg\left(1\right)}=0,2\left(mol\right)\\n_{H_2\left(2\right)}=n_{Fe\left(2\right)}=0,2\left(mol\right)\end{matrix}\right.\)
Ta có: \(n_{H_2}=\Sigma n_{H_2\left(1\right)}+n_{H_2\left(2\right)}=0,2+0,2=0,4\left(mol\right)\)
=> \(V_{H_2\left(đktc\right)}=0,4.22,4=8,96\left(l\right)\)
PTHH:
Mg + H2SO4 \(\rightarrow\) MgSO4 H2
Fe + H2SO4 \(\rightarrow\)FeSO4 + H2
ADCT: n= \(\dfrac{m}{M}\) ta có:
nMg = 4,8/ 24 = 0,2 (mol)
nFe= 11,2/ 56 = 0,2 (mol)
Theo PTHH ta có \(\Sigma\)nH2 = 0,2 + 0,2 = 0,4 (mol)
ADCT : V= 22,4. n ta có:
Thể tích H2 = 22,4. 0,4 = 8,96 (l)
