Chemistry Homework Help?

[Yusuke Urameshi]

I'm sure I appear as an idiot because I keep asking chemistry questions, but meh. I find this thermochemistry stuff challenging. I've got two problems that I'm not quite sure how to do/start. If you get either one or legitimately attempt to answer either one, +rep will be given. If you could explain your steps or what you did, that'd, of course, be super helpful.

#1. Determine the mass of CO2 produced by burning enough carbon (in the form of charcoal) to produce 450kJ of heat.
C(s)+O2(g)→CO2(g) ΔH rxn=−393.5kJ


#2. Titanium reacts with iodine to form titanium(III) iodide, emitting heat.
2Ti(s)+3I2(g)→2TiI3(s) ΔH rxn = -839 kJ

Determine the masses of titanium and iodine that react if 1.80×10^3kJ of heat is emitted by the reaction.

A. Titanium =

B. Iodine =

 

[Prometheus Beta]

1) You are given the enthalpy of reaction (Im assuming its kj/mole), this is the energy released when you produce a single mole of CO2. Now calculate how many moles of CO2 is produced when 450 kj of energy is released instead of 393.5 kj. And then convert the moles into mass.

2) Divide your total energy released by the enthalpy to get half the number of moles of titanium iodide produced (dividing simply by the enthalpy gives the number of the reactions that must occur to produce the total energy, each reaction producing 2 moles of your product). After that, you know the same number of moles of titanium is required and 3/2 the amount of iodine. Same as before.

Hope this helps but I ain't doing your homework for you bro.

Edit: sorry misread 450 kj for some odd reason.

 

[fiend]

1.Co2

Since 44g of CO2 is produced when 393.5 kJ heat is emitted so:-
450/393.5*44=50.316

2.Ti and I2

We know 2mol Ti react with 3mol I2 producing 839kj heat

So
1.80x10^3/893=2.016g

i.e

2x2.016 mols of Ti=48x2x2.016=193.536g

And

3x2.016 mol of I2= 254x3x2.016=1536.192g.

 

[Ldude]

These are just plug n' chug questions. If you're having trouble, it's probably best to go to your teacher for help instead of us.