Audio Visual Question 22

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Emma Scholes 1L
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Joined: Fri Sep 28, 2018 12:18 am

Audio Visual Question 22

Postby Emma Scholes 1L » Mon Oct 01, 2018 11:18 am

According to the following equation 0.750 g of C6H9Cl3 is mixed with 1.000 kg of AgNO3 in a flask of water. A white solid, AgCl, completely precipitates out. What is the mass of AgCl produced?

C6H9Cl3 + 3AgNO3 --> AgCl + C6H9(NO3)3

Molar Mass : C6H9Cl3 (187.50 g/mol), 3AgNO3 (169.88 g/mol), AgCL (143.32 g/mol)

Can someone explain how to solve this?

kamalkolluri
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Re: Audio Visual Question 22

Postby kamalkolluri » Mon Oct 01, 2018 11:43 am

The first thing to notice is that the equation isn't actually balanced. So do that first.
It should come out to C6H9Cl3 + 3AgNO3 --> 3AgCl + C6H9(NO3)3.
After that convert the gram measurements to moles based on molar mass: use grams/Molar Mass. Then find your limiting reactant and use the ratios from the balanced equation to then find the moles of AgCl that precipitates out. After you find the moles then you can use the molar mass and multiply by the moles to get the weight of AgCl that was produced!

Neil Hsu 2A
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Joined: Fri Sep 28, 2018 12:16 am

Re: Audio Visual Question 22

Postby Neil Hsu 2A » Mon Oct 01, 2018 12:03 pm

For this type of problem, you'll want to first balance the chemical equations, so balancing it, you get...

C6H9Cl3 + 3AgNO3 --> 3AgCl + C6H9(NO3)3

Next, convert the grams of reactants into moles. So,

0.750g C6H9Cl3 / 187.50g mol-1 = 0.004 mol C6H9Cl3

1000g AgNO3 / 169.88g mol-1 = 5.886 mol AgNO3
*Btw, the 3AgNO3 (169.88 g/mol) in the problem is a typo since one mole of AgNO3 is 169.88g

Next, you'll want to see which one is the limiting reactant, so comparing calculated moles to required moles, the molar ratio for C6H9Cl3 to AgCl is 1:3 and the molar ratio for AgNO3 to AgCl is 3:3. Comparing the calculated moles to the required moles, we can see that the limiting reactant is C6H9Cl3 since it creates less product relative to AgNO3. Thus, using dimensional analysis, we can figure out the mass of AgCl using the moles of the limiting reactant (C6H9Cl3).

0.004mol C6H9Cl3 x (3mol AgCl / 1mol C6H9Cl3) x (143.32g / mol AgCl) = 1.72 g AgCl
You can make 1.72g AgCl.

Hai-Lin Yeh 1J
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Re: Audio Visual Question 22

Postby Hai-Lin Yeh 1J » Mon Oct 01, 2018 3:18 pm

Step 1: Check if the chemical equation is balanced. In this case, it isn't. Thus, the rewritten, correct chemical equation would be:
C6H9Cl3 + 3AgNO3 -> 3 AgCl +C6H9(NO3)3.

Then, it always helps if you write out what they give you, what they do not give you, and what you are trying to find. In this case, they give you the mass of C6H9Cl3 and AgNO3. What they do not tell you is which reactant is the limiting reactant. (The reason you can't just use any reactant is because the limiting reactant determines how much product you're acutally going to make. In other words, the reactant in short supply LIMITS the quantity of product that can be formed.)
[It's like if you're baking brownies and the recipe states that 100 grams of flour 200 grams of sugar will produce 12 pieces. If you have less than 100 grams of flour or 200 grams of sugar, you won't be able to make 12 pieces. You will make less.]

Known: 0.750 g of C6H9Cl3, 1.000 kg of AgNO3
Unknown: Limiting reactant and mass of AgCL produced.

Step 2: Use molar mass to convert grams of reactants into moles:

0.0750g C6H9Cl3 ( 1 mol / 187.50 g) = 0.004 mol C6H9Cl3
*They gave you the mass of AgNO3 in kg. You need to convert it into grams to use molar mass. (1.000 kg = 1000 g)
1000g AgNO3 ( 1 mol / 169.88 g) = 5.886 mol AgNO3

Step 3: Find out which is the limiting reactant by comparing calculated moles to required moles.
0.004 moles of C6H9Cl3 requires 0.012 moles of AgNO3 (0.004 x 3) [The 3 comes from the stoichiometric coefficient in the chemical equation]. Since there is 5.886 mol of AgNO3 available, that means there is an excess amount of AgNO3 and not enough C6H9Cl3. You can see that 5.886 is greater than 0.012 so you have more than you actually need, but that means you don't have enough of the other reactant. Thus, C6H9Cl3 is your limiting reactant.

Step 4: Use your limiting reactant to find out how many moles of product is produced. Based on the stoichiometric coefficients in the chemical equation, 1 mol of C6H9Cl3 will produce 3 moles of AgCl. In step 2, you found out that you have 0.004 mol of C6H9Cl3. Therefore,
0.004 mol of C6H9Cl3 (3 mol AgCl / 1 mol of C6H9Cl3) = 0.012 mol of AgCl

Step 5: Convert 0.012 mol of AgCl back into grams (mass).
0.012 mol of AgCl (143.32 g / 1 mol ) = 1.72g AgCl.


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