Can I Use Stoichiometry to Predict the Products of a Reaction?

[Lori]

Homework Statement

Homework Equations

M = n/L[/B]

The Attempt at a Solution

My answer is A) 1.02 M but I am not sure if i did this right. Can someone explain why this might be right or wrong?

I got moles of Al(NO3)3 =0.0393 moles
I got moles of Ca(NO3)2 = 0.0426 moles by multiplying liters and molarity

then i added the moles of NO3 in each of them . So for Al(NO3)3 , the mole ration is 1/3 so i get that there is 0.1179 moles NO3 from that compound.

then i added the NO3 from Ca(NO3)2 using the same ratio so i get 0.0852.

I added these moles of NO3 together and divided by total liters of solution and got 1.02 MMy question is am i suppose to treat each compound as it were to break into nitrates and use the mole ratios on each of them to find the amt of nitrates they break down into? I wasn't sure! At first, i thought i had to find a limiting reactant and use only one compound to calculate nitrate.

[/B]

 

[Borek]

Your answer is a correct one.

Limiting reagent is a concept related to the reaction. Is there any reaction taking place?

 

[Lori]

Your answer is a correct one.

Limiting reagent is a concept related to the reaction. Is there any reaction taking place?

Now that you ask. I don't think so because they both stay soluble. None of it precipitated into a solid

 

[Lori]

Now that you ask. I don't think so because they both stay soluble. None of it precipitated into a solid

So both of the compounds formed dissociate right? So we get no3 from both the compounds

 

[Borek]

No reaction, no limiting reagent. You just mix two things together. It happens that they both contain the same ion, so this particular ion amount is a sum of amounts that were present in original solutions. And that's how you solved the problem.

 

[Lori]

No reaction, no limiting reagent. You just mix two things together. It happens that they both contain the same ion, so this particular ion amount is a sum of amounts that were present in original solutions. And that's how you solved the problem.

Is there no reaction because the compounds end up being soluble when o wrote the products?

 

[Borek]

There is no reaction because nothing changes - the ions remain exactly as they were before mixing of the solutions.

 

[Lori]

There is no reaction because nothing changes - the ions remain exactly as they were before mixing of the solutions.

oh. Since the keyword in the problem is "mixing", nothing happens right? Or am i missing something?

 

[Lori]

Ok... after thinking a bit about this more. The solutions when mixed create 2 aqueous solutions. Which means that both dissociate into the ions separately. So the amount of nitrate is just added from both the compounds.

However, if it had created a solid, there is a reaction and that means there would be a limiting reactant because only so much of that solid can be form. What's leftover would be the concentration of nitrate of there was a reaction.

Haha. I'm just writing this down just so I can write down my thoughts!

 

[Borek]

However, if it had created a solid, there is a reaction

Yes, but (in general) possible reactions are not limited to a solid being created. Things in the solution can combine producing another soluble product, they can react producing gas and so on.

 

[Lori]

Yes, but (in general) possible reactions are not limited to a solid being created. Things in the solution can combine producing another soluble product, they can react producing gas and so on.

Wait I'm still confused. Did the reactant go through double displacement then?

 

[Borek]

In this case there was no reaction, period.

What I am referring to is that you listed a precipitate as if producing an insoluble salt was the only possible way of things reacting. This is actually only one of many possible ways.

 

[Lori]

In this case there was no reaction, period.

What I am referring to is that you listed a precipitate as if producing an insoluble salt was the only possible way of things reacting. This is actually only one of many possible ways.

So whenever I predict the products by doing double displacement, I have to check if a reaction occurs? Otherwise they stay in the original form?

 

[Borek]

So whenever I predict the products by doing double displacement, I have to check if a reaction occurs?

Aren't you putting a cart before the horse? How can you predict products of the reaction that doesn't occur?

I guess what you mean here is that you are trying to predict whether a reaction occurs by combining ions and checking whether they can react with each other. That's OK., but checking whether they react is not equivalent of saying they do react.

Otherwise they stay in the original form?

Yes, if they don't react they - quite predictably - stay as they are.

 

[Lori]

Aren't you putting a cart before the horse? How can you predict products of the reaction that doesn't occur?

I guess what you mean here is that you are trying to predict whether a reaction occurs by combining ions and checking whether they can react with each other. That's OK., but checking whether they react is not equivalent of saying they do react.
Yes, if they don't react they - quite predictably - stay as they are.

Thank you. I didn't realize that until you mentioned it. Thanks!