Molarity of Reactants: SO3, NH4Cl, HCl

  • Thread starter Soaring Crane
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In summary, we have discussed the preparation of solutions of sulfuric acid, hydrochloric acid, and the acid-base neutralization reaction involving solid ammonium chloride and solid barium hydroxide. We have also determined the concentration of sulfur in the sulfuric acid solution and the molarity of the hydrochloric acid solution. For the neutralization reaction, we have calculated the amount of solid barium chloride formed. There is also a discussion on the limiting reactant and the use of significant digits. Additional clarification is requested for the amount of solid ammonium chloride.
  • #1
Soaring Crane
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Please check my work!

1) In the preparation of a solution of sulfuric acid, 26.7g of SO3 was dissolved in enough water to prepare 8.20L of solution. What is the concentration of sulfur (moles per liter) in this solution?

Equation given: SO3 (g) + H2O (l) --> H2SO4 (aq), which is already balanced

M = mol of solute/ L of solution

I don't know if this is correct because I am confused about finding the concentration of just sulfur:

26.7 g SO3 (1 mol SO3 / 80.07 g SO3) = 0.33346 mol SO4

Now is it (0.33326 mol)/ 8.20 L = M = 0.406659?



2) The acid-base neutralization reaction below is often used as a demonstration of a "spontaneous" endothermic process. In this reaction solid ammonium chloride reacts with solid barium hydroxide to form barium chloride, ammonia, and water. If 8.0g of solid ammonium chloride are mixed with 62.1g of solid barium hydroxide, how many grams of solid barium chloride are formed? Report your answer to two decimal places.

Equation given: NH4Cl (s) + Ba(OH)2 (s) --> BaCl2 (s) + NH3 + H2O (l)

How I Balanced It: 2NH4Cl (s) + Ba(OH)2 (s) --> BaCl2 (s) + 2NH3 + 2H2O(l)

80 g NH4Cl (1 mol NH4Cl / 53.492 g NH4Cl) = 1.49556 mol NH4Cl

62.1 g Ba(OH)2 (1 mol Ba(OH)2 / 171.346 g Ba(OH)2) = 0.36242 mol Ba(OH)2

Now the limiting reactant is Ba(OH)2 since there is excess NH4Cl ?

0.36242 mol Ba(OH)2 ( 1 mol BaCl2 / 1 mol Ba(OH)2 ) (208.23 g BaCl2 / 1 mol BaCl2) = 75.47 g BaCl2 ?



3) In the preparation of a solution of hydrochloric acid, 13.200g of hydrogen chloride was dissolved in enough water to prepare 5.900L of solution. What is the molarity of this solution?

13.200 g HCl (1 mol / 36.458 g HCl) = 0.36206 mol HCl

0.36206 mol HCl / 5.900 L = 0.06137 M ?

THANK YOU.
 
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  • #2
1. OK (although you abuse significant digits)

2. 8.0g or 80g of NH4Cl?

3. OK
 
  • #3
1. 0.407 M, then?

2. Sorry! It's 8.0 g for NH4Cl.

Here I go again:

8.0 g NH4Cl (1 mol NH4Cl / 53.492) = .1495551 mol NH4Cl

Now NH4Cl is the limiting reactant because it's supposed to have twice the mol of Ba(OH)2 but it does not?

.1495551 mol NH4Cl (1 mol BaCl2 / 2 mol NH4Cl) (208.23 g BaCl2 / 1 mol BaCl2) = 15.57 g BaCl2 ?

Thanks.
 
Last edited:
  • #4
Any volunteers?

Thanks.
 
  • #5
Will anyone confirm if I did #2 correctly with 8 g not 80 g?

Thanks.
 
  • #6
Anyone at all please?

Thanks.
 
  • #7
OK.
 

1. What is the molarity of SO3?

The molarity of SO3, or sulfur trioxide, is the concentration of a solution expressed in moles of SO3 per liter of solution. It can be calculated by dividing the moles of SO3 present by the volume of the solution in liters.

2. How do I calculate the molarity of NH4Cl?

To calculate the molarity of NH4Cl, or ammonium chloride, you need to know the number of moles of NH4Cl and the volume of the solution in liters. Divide the moles of NH4Cl by the volume of the solution in liters to get the molarity.

3. What is the molarity of HCl in a 1.5 M solution?

The molarity of HCl, or hydrochloric acid, in a 1.5 M solution is 1.5 moles of HCl per liter of solution. This means that for every liter of the solution, there are 1.5 moles of HCl present.

4. How does the molarity of reactants affect a chemical reaction?

The molarity of reactants can affect the rate of a chemical reaction. In general, a higher molarity of reactants will result in a faster reaction, as there are more particles present to collide and react with each other. Additionally, the molarity of reactants can also affect the equilibrium position of a reaction.

5. Can the molarity of reactants change during a reaction?

Yes, the molarity of reactants can change during a reaction. As reactants are used up and products are formed, the molarity of the reactants will decrease while the molarity of the products will increase. This is due to the Law of Conservation of Mass, which states that matter cannot be created or destroyed in a chemical reaction.

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