By oxidation of 9 g of glucose it is released 150 kJ of heat. Calculate the standard enthalpy of derivation of glucose (in kJ/mol), if the given enthalpies of the reactants are ΔfH (H2O(l)) = -285,8 kJ/mol; ΔfH (CO2(g)) = -393,5 kJ/mol. Mr (glucose) = 180 (The answer is -1075.8)
$$\ce{6 CO2 + 6 H2O -> C6H12O6 + 6O2}$$
$$n(\ce{C6H12O6})=\frac{\pu{9g}}{\pu{180 g mol^-1}}=\pu{0.05 mol}~\ce{C6H12O6}$$
Now, $$\frac{\pu{0.05 mol}}{\pu{150 kJ}}=\frac{\pu{1 mol}}{x}\implies x=\pu{-3000 kJ/mol}$$
$$\Delta_\mathrm rH=\Delta_\mathrm fH(\ce{H2O(l)})+\Delta_\mathrm fH(\ce{CO2(g)})-\Delta_\mathrm fH(\ce{C6H12O6(s)})$$ $$= 6 \times (\pu{-285.8 kJ/mol}) + 6 \times (\pu{-393.5 kJ/mol}) - (\pu{-3000 kJ/mol}) = \pu{-1075.8 kJ/mol}$$
Did I take the right steps?