\(a,n_{Fe_2O_3}=\dfrac{1,2044.10^{22}}{6,022.10^{23}}=0,02\left(mol\right)\\ b,n_{Mg}=\dfrac{7,5275.10^{24}}{6,022.10^{23}}=12,5\left(mol\right)\)
a, 1,2044.1022 phân tử Fe 2 O 3 bằng 1,2044.1022 /6,022.1023 = 0,02 mol
b, 7,5275.10 24 nguyên tử Mg bằng 7,5275.10 24 / 6,022.10 23 = 12,5 mol