Question

Based on the reaction below, how many grams of iron can be melted by adding 41.40 kJ of energy?

Fe (s) →  Fe (l)     ΔH_fus = + 13.80 kJ mol^-1

Answer 1

$$Q=\frac{m}{M}\Delta H\\41.4=\frac{m}{55.845}(13.8)\\m=167.5\ g$$