# Question #0295b

##### 1 Answer

#### Answer:

Zinc metal will be the limiting reagent.

#### Explanation:

**!! LONG ANSWER !!**

You can determine the **limiting reagent** by using the information you provided here, but in order to find the *enthalpy change of reaction*, *assumptions*.

Let's focus on finding the limiting reagent first. Start by writing the balanced chemical equation that describes this **single replacement reaction**

#color(red)(2)"AgNO"_ (3(aq)) + "Zn"_ ((s)) -> "Zn"("NO"_ 3)_ (2(aq)) + 2"Ag"_((s))#

Silver nitrate, **mole ratio**, which tells you that the reaction consumes **twice as many moles** of silver nitrate than of zinc metal.

Keep this in mind.

Notice that the problem provides you with the *molarity* and *volume* of the silver nitrate solution. Use these to calculate the **number of moles** of silver nitrate present in that solution

#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution")color(white)(a/a)|)))#

**Do not** forget to convert the volume from *cubic centimeters* to *cubic decimeters*

#n_(AgNO_3) = "0.100 mol" color(red)(cancel(color(black)("dm"^(-3)))) * overbrace(50.00 * 10^(-3)color(red)(cancel(color(black)("dm"^3))))^(color(blue)("volume in dm"^3))#

#n_(AgNO_3) = "0.00500 moles AgNO"_3#

Use the **molar mass** of zinc to determine how many *moles* you have in that

#0.200 color(red)(cancel(color(black)("g"))) * "1 mole Zn"/(65.38color(red)(cancel(color(black)("g")))) = "0.00306 moles Zn"#

Now, you know that the reaction consumes **twice as many moles** of silver nitrate than moles of zinc. At this point, you must check to see if you have **enough moles** of silver nitrate to allow for **all the moles** of zinc to take part in the reaction.

That many moles of silver nitrate will allow for

#0.00500 color(red)(cancel(color(black)("moles AgNO"_3))) * "1 mole Zn"/(color(red)(2)color(red)(cancel(color(black)("moles AgNO"_3)))) = "0.00250 moles Zn"#

to take part in the reaction. Since you have *more moles* of zinc than you can use up using the available moles of silver nitrate, you can say that silver nitrate will act as a **limiting reagent**.

In other words, silver nitrate will limit the amount of zinc that can take part in the reaction, i.e. it will be **completely consumed** before all the moles of zinc can react.

#color(green)(|bar(ul(color(white)(a/a)color(black)("AgNO"_3 -> "limiting reagent")color(white)(a/a)|)))#

For the second part of the problem, you will need to *assume* that

thedensityof the resulting solution is equal to#"1.00 g cm"^(-3)# the,specific heatof the resulting solution is equal to the specific heatof water#"4.18 J g"^(-1)""^@"C"^(-1)#

You know that the temperature of the solution **increased** by **exothermic**, i.e. it **gives off heat**.

The idea here is that the heat **given off** by the reaction **is equal** to the heat **absorbed** by the solution.

#color(purple)(|bar(ul(color(white)(a/a)color(black)(q_"absorbed" = - q_"given off")color(white)(a/a)|)))#

This means that the *enthalpy change of reaction* will be **negative**, since you know that a minus sign is used to denote **heat lost**.

Your tool of choice here will be this equation

#color(blue)(|bar(ul(color(white)(a/a)q = m * c * DeltaTcolor(white)(a/a)|)))" "# , where

*change in temperature*, defined as the difference between the **final temperature** and the **initial temperature**

You already know that

#DeltaT = 4.3^@"C" -># the temperatureroseby#4.3^@"C"#

If you take the density of the solution to be equal to

#50.00color(red)(cancel(color(black)("cm"^(-3)))) * "1.00 g"/(1color(red)(cancel(color(black)("cm"^(-3))))) = "50.0 g"#

Plug in your values into the above equation and find the amount of heat **absorbed** by the solution

#q = 50.00 color(red)(cancel(color(black)("g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) * color(red)(cancel(color(black)(""^@"C")))) * 4.3color(red)(cancel(color(black)(""^@"C")))#

#q = "898.7 J"#

Now, you can express the enthalpy change of reaction in *kilojoules per mole of silver nitrate*.

You know that **given off** when **moles** of silver nitrate react. This means that you have

#1 color(red)(cancel(color(black)("mole AgNO"_3))) * (898.7 color(red)(cancel(color(black)("J"))))/(0.00500color(red)(cancel(color(black)("moles AgNO"_3)))) * "1 kJ"/(10^3color(red)(cancel(color(black)("J")))) = "179.7 kJ"#

The enthalpy change of reaction *per mole* of silver nitrate will thus be

#DeltaH_("rxn / mol AgNO"_3) = color(green)(|bar(ul(color(white)(a/a)-"179.7 kJ mol"^(-1)color(white)(a/a)|)))#

Remember, the *minus sign* symbolizes **heat lost**.

I'll leave the answer rounded to four **sig figs**.