Stoichiometry - 1 | Chemical Calculations

From the chemical equation: 12 moles of CO₂ are produced for 2 moles of C₆H₆.
⇒ $\frac{{\mathrm{n}}_{\mathrm{CO}2}}{12}$ = $\frac{{\mathrm{n}}_{\mathrm{C}6\mathrm{H}6}}{2}$ ⇒ n_CO2 = 6 n_C6H6

n_CO2 = 6x $\frac{{\mathrm{N}}_{\mathrm{C}6\mathrm{H}6}}{{\mathrm{N}}_{\mathrm{A}}}$ = 6 x $\frac{56 \times {10}^{15}}{6.022 \times {10}^{23}}$ = 5.581 x 10^-7 mol

m_CO2 = n_CO2 x M_CO2 = 5.581 x 10^-7 x 44.0 = 2.456 x 10^-5 g