If excess hydrochloric acid reacts with 4.6g of barium nitrate

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In summary, the problem is to determine the amount of barium chloride produced when excess hydrochloric acid reacts with 4.6g of barium nitrate. The equation for the reaction is 2HCl + BaN2O6 --> BaCl2 + 2HNO3. Using mole proportions, the amount of HCl needed can be calculated by solving for liters using the formula M = moles/liters and knowing that 2 moles of HCl are needed for each mole of barium nitrate. This will result in the same number of moles of barium chloride, from which the grams can be calculated.
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ussjt
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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.
 
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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.
 
  • #3
ussjt said:
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.
 

FAQ: If excess hydrochloric acid reacts with 4.6g of barium nitrate

What is the chemical equation for the reaction between excess hydrochloric acid and 4.6g of barium nitrate?

The chemical equation is: Ba(NO3)2 + 2HCl → BaCl2 + 2HNO3

How many grams of barium chloride are produced in the reaction?

Using the chemical equation, we can calculate the molar mass of barium chloride (BaCl2) and barium nitrate (Ba(NO3)2). Then, we can use stoichiometry to determine the mass of barium chloride produced. The answer is approximately 3.6g of barium chloride.

How many moles of hydrochloric acid are needed for complete reaction with 4.6g of barium nitrate?

To determine the number of moles of hydrochloric acid needed, we must first convert the mass of barium nitrate to moles using its molar mass. Then, using the mole ratio from the chemical equation, we can calculate the moles of hydrochloric acid needed. The answer is approximately 0.029 moles of HCl.

What is the limiting reagent in this reaction?

The limiting reagent is the reactant that is completely consumed in a chemical reaction. In this case, it is the reactant that is present in the smallest quantity. To determine the limiting reagent, we can compare the number of moles of each reactant and see which one is present in the smallest amount. Whichever reactant has the smallest number of moles is the limiting reagent.

What is the theoretical yield of barium chloride in this reaction?

The theoretical yield is the maximum amount of product that can be obtained from a chemical reaction. To calculate the theoretical yield of barium chloride, we must first determine the limiting reagent. Then, we can use stoichiometry to calculate the maximum amount of barium chloride that can be produced from the given amount of limiting reagent. The answer is approximately 3.6g of barium chloride.

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