`n_(Fe) = 28 / 56 = 0.5 ( mol ) `
`n_(CaCO_3) = 10 / 100 = 0.1 ( mol ) `
`n_(H_2SO_4) = 49 / 98 = 0.5 ( mol ) `
`n_(Al_2(SO_4)_3) = 17.1 / 342 = 0.05 ( mol ) `
`n_(O_2) = 5.6 / 22.4 = 0.25 ( mol ) `
`n_(CO_2) = 1.12 / 22.4 = 0.5 ( mol ) `
`n_(NO_2) = 2.24 / 22.4 = 0.1 ( mol ) `
`n_(CaO) =` $\frac{1.5 * 10^{23}}{6 * 10^{23}}$ `= 0.25 ( mol ) `
`n_(Mg) =` $\frac{0.3 * 10^{23}}{6 * 10^{23}}$ `= 0.05 ( mol ) `
`n_(O_2) =` $\frac{6 * 10^{23}}{6 * 10^{23}}$ `= 1 ( mol ) `
