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Chapter 3: Problem 16

Consider the equation \(2 A+B \longrightarrow A_{2} B .\) If you mix \(1.0 \mathrm{~mol}\) of \(A\) with \(1.0 \mathrm{~mol}\) of \(B\), what amount (moles) of \(A_{2} B\) can be produced?

Short Answer

Expert verified
The limiting reactant in the given reaction is A, since we have a smaller amount of A based on the stoichiometric ratio (2:1). Therefore, using the mole ratio of A to \(A_2B\) (2:1), we can calculate that 0.5 moles of \(A_2B\) can be produced when mixing 1.0 mol of A and 1.0 mol of B.

Step by step solution

01

Identify the limiting reactant

In this reaction, we have 1.0 mole of A and 1.0 mole of B. To identify the limiting reactant, we need to compare the mole ratio of A and B according to the reaction equation. The mole ratio of A to B is 2:1 (from 2A and B in the equation). Now divide the available moles of each reactant by their respective mole ratio: - For A: \( \frac{1.0}{2} = 0.5 \) - For B: \( \frac{1.0}{1} = 1.0 \) The reactant with the smallest result is the limiting reactant. In this case, it is A with a result of 0.5.
02

Calculate the moles of \(A_2B\) produced

Now that we've identified A as the limiting reactant, we can calculate the moles of \(A_2B\) produced using the mole ratio. Since the mole ratio of A to \(A_2B\) is 2:1, we need to multiply the result obtained in step 1 (0.5) by the mole ratio of \(A_2B\) (1 mole) to find the number of moles produced: Moles of \(A_2B\) = 0.5 * 1 = 0.5 moles So, 0.5 moles of \(A_2B\) can be produced when 1.0 mol of A and 1.0 mol of B are mixed together.

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