Question

Question #0295b

Answer 1

Zinc metal will be the limiting reagent.

`DeltaH_("rxn / mol AgNO"_3) = - "179.7 kJ mol"^(-1)`

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, `DeltaH_"rxn"`, you're going to have to make some 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, `"AgNO"_3`, reacts with zinc metal in a `color(red)(2):1` 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-g"` sample

`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

• the density of the resulting solution is equal to `"1.00 g cm"^(-3)`
• the specific heat of the resulting solution is equal to the specific heat of water, `"4.18 J g"^(-1)""^@"C"^(-1)`

You know that the temperature of the solution increased by `4.3^@"C"`, so right from the start you know that the reaction was 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

`q` - the amount of heat gained / lost
`m` - the mass of the sample
`c` - the specific heat of the substance
`DeltaT` - the change in temperature, defined as the difference between the final temperature and the initial temperature

You already know that

`DeltaT = 4.3^@"C" ->` the temperature rose by `4.3^@"C"`

If you take the density of the solution to be equal to `"1.00 g cm"^(-3)`, you can say that the solution has a mass of

`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 `"898.7 J"` of heat are being given off when `0.00500` 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.