Question #b63ca

!! 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 #1:color(red)(2)# mole ratio between magnesium and hydrochloric acid. This tells you that the reaction will 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-g"# sample

#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, #DeltaH_"rxn"#, for your 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 #DeltaH_"rxn"# will carry a 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

#q# - heat lost / gained
#m# - the mass of the sample
#c# - the specific heat of the sample
#DeltaT# - the change in temperature, defined as the difference between the 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 #24.12^@"C"# to #27.10^@"C"#. Unless told otherwise, you can assume that the specific heat of the solution is equal to that of pure water

#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 #1.8%# #"HCl"# by mass, so its density can be approximated as #"1.007 g/mL"# at room temperature.

To keep things simple, you could assume that the density of the solution is #"1.00 g/mL"#. This means that your sample would have a mass of

#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 #0.01051# moles of hydrochloric acid take part in the reaction. The heat 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 reaction gives off #" 237 kJ/mol"#

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