1) \(n_{N_2}=\dfrac{1,8\times10^{23}}{6\times10^{23}}=0,3\left(mol\right)\)
\(\Rightarrow V_{N_2}=0,3\times22,4=6,72\left(l\right)\)
2) \(n_{Fe}=\dfrac{12,6}{56}=0,225\left(mol\right)\)
Số nguyên tử Fe là: \(0,225\times6\times10^{23}=1,35\times10^{23}\left(nguyêntử\right)\)
3) \(n_{O_2}=\dfrac{20}{32}=0,625\left(mol\right)\)
Số phân tử O2 là: \(0,625\times6\times10^{23}=3,75\times10^{23}\left(phântử\right)\)