Chemistry question? Plz help I'm so confused!!?

Question

To 5.00 ml of a 2.48 M solution of HCl with a mass of 5.240g, 1.31 g of Mg is added. before the addition, the temperature was 19 degrees C, after the solution, the final temperature was 50 degrees C. Mg+2HCl ---->H2 +MgCl2

1. what is the qrxn?
2. what is the limiting reagent?
3. what is change of H?

From what I understand qsolution = Csp(soln) *mass (soln)* change of temperature
qrx= - q solution

qrxn/experimental moles= change of H/#moles of balanced equation.

If you could tell the steps used that would help greatly! Thanks so much for you're help!

Answer 1

wow.. complicated problem...



let's start with the concepts then we'll work it out...

a) because HCl and MgCl2 are both very soluble in water, when the reaction is done we'll have an aqueous solution

b) because Cp solution isn't given, and this system is complicated enough that it must be measured (you won't find it in a table somewhere), we get to assume Cp solution = Cp water

c) we use our 6 steps to solving stoichiometry problems that you have memorized by now to find limiting reagent and moles produced

d) we then calculate dH/mole..and put it in terms of dHrxn...



got all that? here we go...



*** a ***

Q = m Cp dT = (5.140g + 1.31g) x 4.184J/gC x (50 - 19 C) = 836.6 J



*** b ***

here are the six steps. if you don't have them down by now... get busy memorizing...



1) write a balanced equation

2) convert everything given to moles

3) determine limiting reagent

4) convert moles limiting reagent to moles other species

5) convert moles to mass. this is theoretical mass aka theoretical yield

6) % yield = actual recovered mass / theoretical mass x 100%



and the idea is this... the coefficients of the balanced equation are in mole ratios. so we use those mole ratios to convert moles of one species to moles of another.. ok?



*** b. 1 ***

1 Mg(s) + 2 HCl(aq) ---> 1 H2(g) + 1 MgCl2(aq).. is balanced...



*** b. 2 ***

5.00 mL HCl solution x (1L / 1000 mL) x (2.48 moles HCl / L) = 0.01240 moles HCl

1.31g Mg x (1 mole Mg / 24.30g Mg) = 0.05391 moles Mg



*** b. 3 ***

this is tricky and it's important that you know how to find limiting reagent. what we do is this.. we pick a reactant from step b.2. either one. and we use the coefficients of the balanced equation from step b.1 to calculate how much of the other reactant we actually need. Then we compare that amount to the amount we actually have available in step b.2. If we have more than we need, that reactant is in XS (eXcesS). if less, that reactant is limiting. ok?



let's pick HCl and calculate how much Mg we need...

0.01240 moles HCl x (1 mole Mg / 2 moles HCl) = 0.006200 moles Mg

since we have more than that, (we have 0.05391 moles Mg available), Mg is in XS and HCl is limiting the reaction.



and yes, we could have picked Mg to start with...

0.05391 moles Mg x (2 moles HCl / 1 mole Mg) = 0.1078 moles HCl

we only have 0.01240 moles HCl available so HCl is limiting...



see how that works?



*** b. 4 ***

from the balanced equation, 2 moles HCl ---> 1 mole H2

and 2 moles HCl ---> 1 mole MgCl2



so...

0.01240 moles HCl x (1 mole H2 / 2 moles HCl) = 0.006200 moles HCl

0.01240 moles HCl x (1 mole MgCl2 / 2 moles HCl) = 0.006200 moles MgCl2



*** b. 5 ***

and we can convert that all of that to masses...



mass Mg consumed = 0.006200 moles x (24.30g Mg / mole Mg) = 0.1507g

mass H2 produced = 0.006200 moles H2 x (2.016g H2 / mole H2) = 0.01250g

mass MgCl2 produced = 0.006200 moles MgCl2 x (95.20g MgCl2 / mole MgCl2) = 0.5902g



************

and we'll skip step 6...



so this ended up with...a solution containing....

1.31g - 0.15g = 1.16g solid Mg

5.240g - 1.16g - 0.01250g = 4.068g aqueous MgCl2



so if we assume both the solution and the metal are warming then maybe we can get a better estimate of Q...

Q = (m Cp dT)solution + (m Cp dT) metal

Q = (4.068g x 4.184J/gC x 31C) + (1.16g x 1.023 J/gC x 31C) = 564.4 J



that's quite a bit different isn't it? and that is because not all of the Mg metal was consumed. If Mg was the limiting reagent, then Q = 836.6J.. ok?



*** c ***

and finally...that is based on 0.01240 moles of HCl...



so.. dH/mole (HCl) = 564.4J / 0.1240 moles = 45.5 kJ

and dH for 2 moles HCl = 91.0 kJ.. right?



and....

1 Mg(s) + 2 HCl(aq) ---> 1 H2(g) + 1 MgCl2(aq)... dHrxn = - 91.0 kJ

Answer 2

Hint; 1.First check if you reaction is balanced. I dont understand qrxn whether its a formula or abbr. but determining the limiting reagant is very easy.

First calculate the number of moles of HCl= n(HCl)=MM/m

Then calculate the number of moles of Mg =n(Mg) =MM/m



Then the compound between Mg and HCl that has the lowest number of moles will be your limiting reagant.

Answer 3

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