Please help me on this Chemistry question.

Question

20 grams of H₂S react with 20 grams of O₂. What is the limiting reagent?

 

a. H₂S

b. O₂

c. SO₂

d. H₂O

6 moles of O₂ react. Assume you have more than enough H₂S to react with it. How many moles of SO₂ are produced?

 

a. 1 mole

b. 2 moles

c. 4 moles

d. 6 moles

e. Cannot be determined

Answer 1

Hello, Taylor,

The table below summarizes the steps. Determine the molar mass (g/mole) of both reactants and then the number of moles of each, by dividing the grams by the g/mole (grams cancel and moles moves to the top, leaving just moles). The balanced equation tells us we need 3 moles of O₂ for every 2 moles of H₂S. That is a ratio of (3 moles O₂/2 moles H₂S), or 1.5. Multiply the moles H2S by 1.50 to determine how many moles of O₂ are needed (1.50 moles O₂/ moles H₂S) x 0.588 moles H₂S to give 0.882 moles of O₂ that would be required to react with all of the H₂S. We can see that we don't have that many moles of O₂, therefore it is the limiting reagent. H₂S will react with all the available O₂ before the O₂ is depleted, leaving excess H₂S.

For bonus points, we can determine the amount of H₂S remaining by assuming all the O₂ is consumed. the total H₂S consumed would be:

(2 moles H₂S)/(3 moles O₂)*(0.625 moles O₂) = 0.417 moles H₂S.

This would leave 0.208 moles H₂S unreacted. Multiply that by the molar mass of H₂S to get grams of H₂S.

(0.417 moles H₂S)*(34 grams H₂S/mole H₂S) = 7.1 grams.

I hope you can use this reasoning to solve the other problems.

Bob

Answer 2

Hi Taylor H.,

So you are working with two different, but related, measures of matter: 1) mass and 2) moles .

I assume you have the idea of mass, intuitively: it's NOT how much an object weighs, but rather how difficult it is to accelerate it (this includes stopping it too).

Moles, though -- that's a unit that depends on how many individual particles (molecules, or atoms if that is what they are) there are.

When you write a balanced chemical reaction, the coefficients on each material (that's the number out in front of the molecular formula) are talking about moles.

So when you are given problems such as these, you must convert any masses into moles (divide by the molecular weight == formula weight == molar mass, etc. of the material), then compare those results as ratios to the ratios in the balanced reaction. Find out what is in excess (as moles), and only allow the "proper" amount to react, as dictated by the limiting material.

So, in the first reaction, 20g H2S => 20/34.082g/mol => 0.5868 moles => 0.5868/2 = 0.2634 of stated reaction amount of moles.

And 20g O2 => 20/31.9988g/mol => 0.6250 moles => 0.6250/3 => 0.2083 of stated reaction amount of moles.

So the O2 is limiting. If you wanted to find out how much SO2 and H2O could be produced, you would carry over this factor into the products, multiply by 2 for each of the products (since those are the coefficients of each in the reaction), and, if required, convert each to a mass basis using the respective molecular weights.

You should practice this conversion stuff a lot, until you get proficient at it, since you will be doing it a lot in your chemistry course. Such as for the second part of the question!

--Cheers, --Mr. d.