What is the standard enthalpy change for the formation of one mole of strontium carbonate, srco3 (s), from elements?

Answer: (A) Sr (s) + (1/2) O2 (g) - - - > SrO (s) dH = - 592 kJ (B) SrO (s) + CO2 (g) - - - > SrCO 3 (s) dH = - 234 kJ (C) C (graphite) + O2 (g) - - - > CO 2 (g) dH = - 394 kJ to get : Sr (s) + (3/2) O2 (g) + C (graphite) - - - > SrCO 3 (s) we use 1 equation (B) to provide the product, SrCO 3 (s) we use 1 equation (A) to provide the Sr as a starting material we use 1 equation (C) to provide the C as a starting material (we are not concerned with providing the necessary O2, since it is provided from more than 1 source combining 1 (A) & 1 (B) & 1 (C) gives: Sr + (1/2) O2 & SrO (s) + CO2 (g) & C (graphite) + O2 (g) - - - > SrO (s) & SrCO 3 (s) & CO 2 we notice that SrO (s) - - > cancels out.--> SrO (s) CO2 (g) - - > cancels out.--> CO2 (g) leavng only your desired equation Sr (s) + (3/2) O2 (g) + C (graphite) - - > SrCO 3 (s) Hess's law says that if adding 1 (A) & 1 (B) & 1 (C) gives you the equation you wish, the combining the dH's from 1 (A) & 1 (B) & 1 (C) gives you the dH you wish - 592 kJ & - 234 kJ & - 394 kJ = - 1220 kJ your answer is dH = - 1220 kJ