# Question c845f

Apr 11, 2017

$\Delta {H}_{\text{rxn"^@ = - "108 kJ}}$

#### Explanation:

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

$\text{Zn"_ ((s)) + 1/2"O"_ (2(g)) -> "ZnO"_ ((s)) + "350.5 kJ}$

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

$\Delta {H}^{\circ} = - {\text{350.5 kJ mol}}^{- 1}$

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

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\Delta {H}_{\text{rxn"^@ = - "108 kJ}}}}}$

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.