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Chapter 3: Problem 74

Aluminum reacts with sulfur gas to form aluminum sulfide. Initially, \(1.18\) mol of aluminum and \(2.25\) mol of sulfur are combined. (a) Write a balanced equation for the reaction. (b) What is the limiting reactant? (c) What is the theoretical yield of aluminum sulfide in moles? (d) How many moles of excess reactant remain unreacted?

Short Answer

Expert verified
Answer: The theoretical yield of aluminum sulfide is 0.59 moles, and there are 0.48 moles of sulfur remaining unreacted.

Step by step solution

01

(a) Write a balanced equation for the reaction

To write a balanced equation, we need to identify the reactants and products and ensure that the number of atoms of each element is the same on both sides of the equation. In this case, the reactants are aluminum (Al) and sulfur (S), and the product is aluminum sulfide (Al2S3). Balancing the equation gives us: $$ 2\text{Al} + 3\text{S} \to \text{Al}_2\text{S}_3 $$
02

(b) Identify the limiting reactant

A limiting reactant is the reactant that is completely consumed in the reaction and determines the amount of product formed. To find the limiting reactant, compare the ratio of moles of each reactant to the balanced equation coefficients. Mole ratio of aluminum: \(\frac{1.18\,\text{mol}\,\text{Al}}{2} = 0.59\) Mole ratio of sulfur: \(\frac{2.25\,\text{mol}\,\text{S}}{3} = 0.75\) Since the mole ratio of aluminum is lower than that of sulfur, aluminum is the limiting reactant.
03

(c) Calculate the theoretical yield of aluminum sulfide in moles

The theoretical yield is the amount of product that would be formed if the limiting reactant is completely consumed. From the balanced equation, we can see that 2 moles of aluminum reacts with 3 moles of sulfur to form 1 mole of aluminum sulfide. Using the limiting reactant (aluminum) to find the theoretical yield: $$ 1.18\,\text{mol}\,\text{Al} \times \frac{1\,\text{mol}\,\text{Al}_2\text{S}_3}{2\,\text{mol}\,\text{Al}} = 0.59\,\text{mol}\,\text{Al}_2\text{S}_3 $$ So, the theoretical yield of aluminum sulfide is \(0.59\) moles.
04

(d) Calculate moles of excess reactant remaining unreacted

To find the moles of excess reactant remaining unreacted, first, determine how much of the excess reactant was consumed in the reaction and then subtract that amount from the initial moles given. Since aluminum is the limiting reactant, sulfur is the excess reactant. From the balanced equation, 3 moles of sulfur react with 2 moles of aluminum. So, $$ 1.18\,\text{mol}\,\text{Al} \times \frac{3\,\text{mol}\,\text{S}}{2\,\text{mol}\,\text{Al}} = 1.77\,\text{mol}\,\text{S} $$ Now, subtract the moles of sulfur consumed from the initial moles given: $$ 2.25\,\text{mol}\,\text{S} - 1.77\,\text{mol}\,\text{S} = 0.48\,\text{mol}\,\text{S} $$ So, \(0.48\) moles of sulfur remain unreacted.

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