Homework questions thread

[Hamdog17]

[CuCl2]=[Cu2+]=2M
So for every mole of CuCl2, there are 2 moles of Cl- ions.
:. [Cl-]=2*[Cu2+]
    [Cl-]=2*2=4M
   

oh.my.gosh i didn't know it was that simple.

thanks!

You're welcome

[chris19021]

this might be the easiest question posted..i dont quite grasp the concept of finding the amount in excess..
Q: 18.0g of Mg is added to 200mL of 2.00M HCl. The mass in gram, of Mg that is left behind after the reaction is closest to:
a) 4.2
b) 8.4
c) 13.2
d) 15.6

P.S sorry of its a stupid question

[thushan]

OK so how much Mg is there?
Ans = 18.0/24.3 = 0.74 mol

How much HCl?
Ans = cV = 2.00 x 0.200 = 0.40 mol

Now, it takes 2 mol HCl to react with 1 mol of Mg.
Hence it takes 0.40 mol HCl to react with 0.20 mol Mg. (use this thought process to work out excess)

Hence, all the HCl is reacted, and only 0.20 mol of Mg has reacted! But we have 0.74 mol Mg originally.
Hence, the Mg left is 0.74 - 0.20 = 0.54 mol - this is the EXCESS.
Hence m(Mg)excess = 0.54 x 24.3 = 13.1 ~ 13.2 g.

Hence C.

[chris19021]

thanks a lot thushan...it makes a lot of sense now...thanks

[chris19021]

one more question PLEASE:

Q: 1L of a certain hydrocarbon was burnt in excess air at 500C and 1 atm pressure. The CO2 and steam produced in this reaction were found to occupy 5L and 6L respectively when measured at 500C and 1 atm pressure. What was the molecular structure?
a) C₃H₈
b) C₅H₆
c) C₅H₁₂
d) C₆H₁₂

THANKS!!

[Aurelian]

C?H? + XO2 ---> YCO2 + ZH20

PV = nRT
n = PV/RT

n(CO2) = (101.3*5)/(8.31*773) = 0.0788mol
n(H2O) = (101.3*6)/(8.31*773) = 0.0946mol

From the reaction, we can see that all the carbon in the hydrocarbon is contained within the produced CO2, and all the hydrogen is contained within the steam.

Thus;
n(C) = n(CO2) = 0.0788mol
n(H) = 2 x n(H20) = 0.189mol since n(H):n(H20) = 1:2

n(C):n(H) = 1:2.4 = 5:12
Therefore, the molecular structure was C5H12 - answer c).

Hope that helps!

[chris19021]

n(C):n(H) = 1:2.4 = 5:12
Therefore, the molecular structure was C5H12 - answer c).

i dont see where u got the 5 and 12 from..sorry could u explain it.

[Aurelian]

n(C):n(H) = 1:2.4 = 5:12
Therefore, the molecular structure was C5H12 - answer c).

i dont see where u got the 5 and 12 from..sorry could u explain it.

Sure thing

So we get the ratio of 1:2.4, but obviously this as it stands is a nonsense ratio - you need whole numbers here. You can't have one carbon atom boned with "2.4" hydrogen atoms; it just doesn't make sense. What this ratio is is just the 'simplest' representation of what's going on. So we need to get to whole numbers.

So, we scale the ratio by a whole number factor to get a completely whole number ratio. In this case, the factor is 5.

1*5 = 5
2.4*5 = 12

So the ratio 1:2.4 is equivalent to 5:12

Does that help?

[chris19021]

tht makes soo much sense now..thanks alot!!:)

[soccerboi]

I need help with this, i have no idea where to start

Q) When solid lead dioxide,PbO2, is heated, it forms solid lead monoxide, PbO, and oxygen. Heating solid barium peroxide, BaO2, yields solid barium monoxide, BaO, and oxygen. A mixture of lead dioxide and barium peroxide was heated until both decompositions were complete. If the initial mass of the mixture was 15g and the final mass was 13.8g, what mass of lead dioxide was present in the original mixture? Ans is 7.873g

[Panicmode]

I need help with this, i have no idea where to start

Q) When solid lead dioxide,PbO2, is heated, it forms solid lead monoxide, PbO, and oxygen. Heating solid barium peroxide, BaO2, yields solid barium monoxide, BaO, and oxygen. A mixture of lead dioxide and barium peroxide was heated until both decompositions were complete. If the initial mass of the mixture was 15g and the final mass was 13.8g, what mass of lead dioxide was present in the original mixture? Ans is 7.873g

See my attached scan for working

STEPS:
Start with writing the two balanced equations
Write equations for the initial and final mass.
Find the ratios of the molar masses for each of the respective equations
Rearrange and substitute the ratios of the molar masses into the initial mass (or final mass if you want to do it that way) equation
Solve the equations simultaneously.
Use mole ratios to calculate the amount of PbO2
Calculate the mass of PbO2.
 

Hope I helped

[soccerboi]

thanks panicmode, much appreciated (:

[soccerboi]

could u please explain the 3rd equation with the molar mass? I dont understand y and how u got it? Thanks

[Panicmode]

could u please explain the 3rd equation with the molar mass? I dont understand y and how u got it? Thanks

Sure .

The ratios I made were based on the ratios of the masses of each of the substances. To show exactly what I did, here is a step process:

x mole of PbO2 weighs y amount.
x mole of PbO weighs z amount.

Therefore, the mass ratio PbO2 : PbO is y : z.

Now, x could be any amount of mole but to make things simple, I made it equal 1. As we know, one mole of a substance equals the molar mass. Therefore the ratio of m(PbO2)/m(PbO) = M(PbO2)/M(PbO) Then by rearranging the equation, I am able to get m(PbO2) [or m(PbO)] by itself. I then did the same thing to BaO2 and BaO.

Next, I took the equation for initial mass [m(PbO2) + m(BaO2) = 15g] and substituted in my rearranged equation of the mass ratios. This meant that the variables were the same in both equations and so could be solved simultaneously. If you wanted, you could rearrange the mass ratio equations so that m(PbO) and m(BaO) are by themselves. I could then substitute this into the equation for final mass and  it would still work. The aim is to get only two variables.



This is a slightly convoluted explanation so if you need more help feel free to PM me and I'll see if I can explain it another way.

Hope I helped.

[Aurelian]

^ I think it's worth noting that you would never get a stoichiometric question of this kind of difficulty on any VCAA exam, so while it's nice knowing how to do it, don't fret too much if you can't =)