If excess hydrochloric acid reacts with 4.6g of barium nitrate

[ussjt]

First, if this is in the wrong section...sorry. I thought this would be the right place because it is dealing with homework.

Here is the problem:

If excess hydrochloric acid reacts with 4.6g of barium nitrate, how man grams of barium chloride are produced? If the hydrochloric acid is .3M, how much is needed to completely react with the 4.6g of barium nitrate?

I found the equation to be:
2HCL + BaN2O6 --> BaCl2 + 2HNO3

then for the first part I found the answer to the first part to be about 3.67g (if this is wrong please tell me and give a hint to where I messed up).

Now my main problem is that I don't remember how to set up the last part...If you could get me started and provide some hints, that would be great.

Thanks a lot.

 

[Tide]

First figure out how many moles of barium nitrate you start with. Once you have that then you will end up with exactly the same number of moles of barium chloride from which you can calculate the number of grams. Also, to determine the amount of HCl required just observe that two moles of HCl are required for each mole of barium nitrate. It's just ratios and proportions.

 

[needhelpperson]

First, if this is in the wrong section...sorry. I thought this would be the right place because it is dealing with homework.

Here is the problem:

If excess hydrochloric acid reacts with 4.6g of barium nitrate, how man grams of barium chloride are produced? If the hydrochloric acid is .3M, how much is needed to completely react with the 4.6g of barium nitrate?

I found the equation to be:
2HCL + BaN2O6 --> BaCl2 + 2HNO3

then for the first part I found the answer to the first part to be about 3.67g (if this is wrong please tell me and give a hint to where I messed up).

Now my main problem is that I don't remember how to set up the last part...If you could get me started and provide some hints, that would be great.

Thanks a lot.

For the last part remember that M = mols/liters

i'm assuming you know about the mole proportions... once you know that

.3M = (your mole)/ liters ---> solve for liters and that's the answer as to how much of the HCl is needed.