Where Did I Go Wrong in My Concentration Problem?

In summary, the conversation is about a problem involving the production of lithopone, a white pigment used in water-based interior paints. The problem requires determining the mass of lithopone produced in a reaction between 227 mL of 0.257 M ZnSO4 and 325 mL of 0.280 M BaS. The solution involves finding the limiting reagent, calculating the moles of the precipitated solids, and adding their masses to find the final answer.
  • #1
geffman1
67
0

Homework Statement


hey guys I've got a problem, could you tell us where i went wrong

Lithopone is a brilliant white pigment used in water-based interior paints. It is a mixture of barium sulfate and zinc sulfide produced by the reaction
BaS(aq) + ZnSO4(aq) ZnS(s) + BaSO4(s)
What mass of lithopone in grams (to 3 significant figures) is produced in the reaction of 227 mL of 0.257 M ZnSO4 and 325 mL of 0.280 M BaS?

MY WORKED SOLUTION"
Mole of BaS=0.280*0.325=0.091mole
mole of ZnSO4=0.0257*0.277=0.0712moles <therefore this is limiting

i don't know what i should take the molarmass of to work out the mass?

ANSWER IS 1.93 grams

anyhelp would be GREATFULLY appreciated.



Homework Equations





The Attempt at a Solution

 
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  • #2
geffman1 said:
mole of ZnSO4=0.0257*0.277=0.0712moles <therefore this is limiting

Check your math. Check it twice, as abvious typo is not the error I am thinking about.

ANSWER IS 1.93 grams

No, 1.93g is not the answer. Probably typo again.

You are on the right track. Find limiting reagent, find moles of both solids that precipitated, find their masses - and as you are looking for a precipitate mass, add them.
 
  • #3


First, let's determine the limiting reactant in the reaction. We can do this by comparing the moles of each reactant present.

Moles of BaS = 0.280 M * 0.325 L = 0.091 mol
Moles of ZnSO4 = 0.257 M * 0.227 L = 0.058 mol

Since the reaction requires a 1:1 ratio of BaS to ZnSO4, we can see that ZnSO4 is the limiting reactant as it produces less moles of product.

Next, we can use the balanced equation to determine the theoretical yield of lithopone, which is the maximum amount of product that can be produced from the given reactants.

1 mol ZnSO4 produces 1 mol lithopone
0.058 mol ZnSO4 produces 0.058 mol lithopone

Now, we can use the molar mass of lithopone (330.4 g/mol) to convert moles to grams.

0.058 mol lithopone * 330.4 g/mol = 19.18 g lithopone

However, since the question asks for the mass in grams to 3 significant figures, we round the answer to 1.93 grams.

Therefore, 1.93 grams of lithopone is produced in the reaction of 227 mL of 0.257 M ZnSO4 and 325 mL of 0.280 M BaS.

Hope this helps!
 

FAQ: Where Did I Go Wrong in My Concentration Problem?

What is a concentration problem?

A concentration problem is a type of mathematical problem that involves finding the amount or percentage of a substance in a solution. It typically involves using a formula to calculate the concentration of a solution based on its volume and the amount of the substance present.

How do you solve a concentration problem?

To solve a concentration problem, you will need to gather all the necessary information, including the volume of the solution, the amount of the substance present, and any other relevant data. Then, you can use the appropriate formula, such as the molarity formula, to calculate the concentration.

What is the difference between molarity and molality?

Molarity is a measure of the concentration of a substance in a solution, expressed as the number of moles of the substance per liter of solution. Molality, on the other hand, is a measure of the concentration of a substance in a solution, expressed as the number of moles of the substance per kilogram of solvent. Molarity takes into account the volume of the solution, while molality takes into account the mass of the solvent.

How do you convert between molarity and molality?

To convert between molarity and molality, you will need to know the density of the solution. Then, you can use the formula: molality = molarity / (density in g/mL). This will allow you to convert from molarity to molality, or vice versa.

What are some real-world applications of concentration problems?

Concentration problems have many practical applications in fields such as chemistry, biology, and medicine. For example, they can be used to determine the strength of a medication, the amount of a particular substance in a sample, or the concentration of pollutants in a water source. They are also commonly used in industries such as food and beverage production to ensure the correct amount of ingredients are present in a solution.

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