# Question #b63ca

##### 1 Answer

#### Answer:

#### Explanation:

**!! LONG ANSWER !!**

The first thing to do here is write a balanced chemical equation for this single replacement reaction

#"Mg"_text((s]) + color(red)(2)"HCl"_text((aq]) -> "MgCl"_text(2(aq]) + "H"_text(2(g]) uarr#

Notice that you have a **always** consume **twice as many moles** of hydrochloric acid than you have moles of magnesium that *take part in the reaction*.

Use magnesium's molar mass to determine how many moles you have in that

#0.1277 color(red)(cancel(color(black)("g"))) " 1 mole Mg"/(24.3050color(red)(cancel(color(black)("g")))) = "0.005254 moles Mg"#

Now use the molarity and volume of the hydrochloric acid solution to figure out how many moles of acid you have

#color(blue)(c = n/V implies n = c * V)#

#n_(HCl) = "0.500 M" * 200.0 * 10^(-3)"L" = "0.100 moles HCl"#

Since you have more moles of hydrochloric acid than would have been required for all the moles of magnesium to react, magnesium will act as a limiting reagent.

This means that **all the moles** of magnesium will take part in the reaction. Moreover, the reaction will consume

#0.005254 color(red)(cancel(color(black)("moles Mg"))) * (color(red)(2)" moles HCl")/(1color(red)(cancel(color(black)("mole Mg")))) = "0.01051 moles HCl"#

Keep this in mind.

Now focus on the finding the **enthalpy change of reaction**,

Notice that performing this reaction leads to an **increase** in the temperature of the solution. This tells you that the reaction is **giving off heat** to the solution.

This implies that *negative sign*.

Now, the relationship between heat lost / gained and change in temperature is described by the following equation

#color(blue)(q = m * c * DeltaT)" "# , where

*final temperature* and the *initial temperature* of the sample

Use this equation to determine how much heat was needed to raise the temperature of the solution from

#c = 4.18"J"/("g" ""^@"C")#

Now, the density of the hydrochloric acid solution can be found here

http://www.handymath.com/cgi-bin/hcltble3.cgi?submit=Entry

Your solution is about

To keep things simple, you *could* assume that the density of the solution is

#200.0 color(red)(cancel(color(black)("mL"))) * "1.00 g"/(1.00color(red)(cancel(color(black)("mL")))) = "200.0 g"#

This means that the solution absorbed

#q = 200.0 color(red)(cancel(color(black)("g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * (27.10 - 24.12)color(red)(cancel(color(black)(""^@"C")))#

#q = "2491.3 J"#

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

#DeltaH_"rxn" = - q#

The *minus sign* is used because **heat lost** carries a negative sign.

This means that the enthalpy change for this reaction is

#DeltaH_"rxn" = -"2491.3 J"#

Now, this is how much heat is given off when **one mole** of hydrochloric acid reacts will be

#1 color(red)(cancel(color(black)("mole HCl"))) * "2491.3 J"/(0.01051color(red)(cancel(color(black)("moles HCl")))) = "237, 041 J"#

Rounded to three sig figs, the number of sig figs you have for the molarity of the hydrochloric acid solution, the answer will be

#DeltaH_"rxn" = color(green)(-"237 kJ/mol")#

Remember, enthalpy change **must** carry a negative sign when heat is being **given off**!

These two statements are **equivalent**

The reactiongives off#" 237 kJ/mol"#

The enthalpy change of reaction is#-"237 kJ/mol"#