Consider the hypothetical reaction between A2 and AB pictured below. (picture of eight AB molecules and four A2 molecules in one box -> eight molecules of A2B molecules in another box) What is the balanced equation? If 2.50 mol A2 is reacted with excess AB, what amount (moles) of product will form? If the mass of AB is 30.0 amu and the mass of A2 is 40.0 amu, what is the mass of the product? If 15.0 g of AB is reacted, what mass of A2 is required to react with all of the AB, and what mass of product is formed?

Question

Do you have some idea of what is to be done? I don't want to work all of the problem for you. Perhaps you are stuck on just one part. Here is the reaction, as I see it.
A2 + 2AB ==> 2A2B

Please post with specific questions and tell me what you don't understand. I can help you better that way.

Answer 1

Sure! Let's break down the problem into parts.

1) Balanced equation: A2 + 2AB ==> 2A2B (as you mentioned)

2) Given 2.50 mol A2, calculate moles of product formed:
Since the balanced equation shows a 1:2 ratio between A2 and A2B, we can multiply the moles of A2 by 2 to find the moles of product formed:
2.50 mol A2 * 2 mol A2B/mol A2 = 5.00 mol A2B

3) Calculate the mass of product:
First, we need to find the molar mass of A2B. We're given that the mass of AB is 30.0 amu and the mass of A2 is 40.0 amu (so the mass of A is 20.0 amu since there are two A atoms in A2). Thus, the mass of A2B is 20.0 amu + 30.0 amu = 50.0 amu. Now, we can find the mass of the product by multiplying moles of A2B by the molar mass:
5.00 mol A2B * 50.0 g/mol A2B = 250.0 g A2B

4) Given 15.0 g of AB, calculate mass of A2 required and mass of product formed:
First, we need to find the moles of AB. Using the molar mass of AB (30.0 g/mol), we get:
15.0 g AB * (1 mol AB / 30.0 g AB) = 0.50 mol AB

Now, we can use the balanced equation to find the required moles of A2 (1:2 ratio between A2 and AB):
0.50 mol AB * (1 mol A2 / 2 mol AB) = 0.25 mol A2

Convert moles of A2 to mass using the molar mass of A2 (40.0 g/mol):
0.25 mol A2 * 40.0 g/mol A2 = 10.0 g A2

To find the mass of the product formed, we can use the balanced equation again (1:2 ratio between A2 and A2B):
0.25 mol A2 * 2 mol A2B/mol A2 = 0.50 mol A2B

Finally, convert moles of A2B to mass using the molar mass of A2B (50.0 g/mol):
0.50 mol A2B * 50.0 g/mol A2B = 25.0 g A2B

In summary:
- 5.00 moles of A2B will form when 2.50 mol A2 is reacted with excess AB
- The mass of the product (A2B) is 250.0 grams
- 10.0 grams of A2 is required to react with all of the AB (15.0 grams), and 25.0 grams of A2B will be formed