Question #c845f

You are definitely on the right track here.

You know that you are dealing with an exothermic reaction because heat is added as a product

#"Zn"_ ((s)) + 1/2"O"_ (2(g)) -> "ZnO"_ ((s)) + "350.5 kJ"#

This tells you that when #1# mole of zinc reacts with #1/2# moles of oxygen gas and #1# mole of zinc oxide is produced, #"350.5 kJ"# of heat are being given off. You can thus say that the standard molar enthalpy change for this reaction is equal to

#DeltaH^@ = -"350.5 kJ mol"^(-1)#

The minus sign is used because heat is being given off by the reaction.

Use the molar mass of zinc oxide to convert the mass to moles

#25.0 color(red)(cancel(color(black)("g"))) * "1 mole ZnO"/(81.408color(red)(cancel(color(black)("g")))) = "0.3071 moles ZnO"#

Now, you can use the standard molar enthalpy change for this reaction as a conversion factor to help you figure out the standard enthalpy change of reaction that occurs when #0.3071# moles of zinc oxide are produced

#0.3071 color(red)(cancel(color(black)("moles ZnO"))) * overbrace((-"350.5 kJ")/(1color(red)(cancel(color(black)("mole ZnO")))))^(color(blue)(= - "350.5 kJ mol"^(-1))) = -"108 kJ"#

You can thus say that when #0.3071# moles of zinc oxide are being produced, the standard wenthalpy change of reaction is equal to

#color(darkgreen)(ul(color(black)(DeltaH_"rxn"^@ = - "108 kJ")))#

The answer is rounded to three sig figs, the number of sig figs you have for the mass of zinc oxide produced by the reaction.