Answer:
17.3 g
Step-by-step explanation:
Let's consider the combustion of CO.
CO(g) + 0.5 O₂(g) → CO₂(g)
We can find the standard enthalpy of the reaction using the following expression.
ΔH°rxn = 1 mol × ΔH°f(CO₂(g)) - 1 mol × ΔH°f(CO(g)) - 0.5 mol × ΔH°f(O₂(g))
ΔH°rxn = 1 mol × (-393.5 kJ/mol) - 1 mol × (-110.5 kJ/mol) - 0.5 mol × 0
ΔH°rxn = -283.0 kJ
283.0 kJ are released per mole of CO. The moles of CO required to release 175 kJ are:
(-175 kJ) × (1 mol CO/ -283.0 kJ) = 0.618 mol CO
The molar mass of CO is 28.01 g/mol. The mass of CO is:
0.618 mol × (28.01 g/mol) = 17.3 g