Given the equation below, what is D H for burning 10.0 g of CO (g) in excess oxygen?
2 CO (g) + O₂ (g) →2 CO₂ (g) D H = - 566 kJ

Question

Given the equation below, what is D H for burning 10.0 g of CO (g) in excess oxygen? 2 CO (g) + O₂ (g)→2 CO₂ (g) D H = - 566 kJ

Quizplus · Accepted Answer

The balanced equation shows that 2 moles of CO react with 1 mole of O2 to produce 2 moles of CO2. The enthalpy change for this reaction is -566 kJ. To find the enthalpy change for burning 10.0 g of CO, we need to convert the mass of CO to moles and then use the mole ratio from the balanced equation to find the moles of O2 that react.10.0 g CO x (1 mol CO / 28.01 g CO) = 0.357 mol COFrom the balanced equation, we see that 2 moles of CO react with 1 mole of O2. Therefore, 0.357 mol CO will react with 0.1785 mol O2.The enthalpy change for burning 10.0 g of CO is:-566 kJ / 2 mol CO x 0.357 mol CO = -101 kJTherefore, the correct answer is A) 101 kJ.

Given the Equation Below, What Is D H for Burning