Answer :

100.2 grams of aluminum sulfide can be produced wher 2.00 mol of aluminum react with 2.00 mol of sulfur.

To determine how many grams of aluminum sulfide[tex](\(\text{Al}_2\text{S}_3\))[/tex] can be produced from 2.00 moles of aluminum and 2.00 moles of sulfur, follow these steps:

Step 1: Write the Balanced Chemical Equation

  • The balanced chemical equation for the reaction between aluminum and sulfur to produce aluminum sulfide is:
  • [tex]\[ 2 \text{Al} (s) + 3 \text{S} (s) \rightarrow \text{Al}_2\text{S}_3 (s) \][/tex]

Step 2: Determine the Limiting Reactant

  • From the balanced equation:
  • 2 moles of aluminum react with 3 moles of sulfur to produce 1 mole of [tex]\(\text{Al}_2\text{S}_3\).[/tex]

Given:

  • 2.00 moles of aluminum
  • 2.00 moles of sulfur

Calculate the required moles of sulfur for the available moles of aluminum:

[tex]\[ \text{Moles of sulfur needed} = \frac{3 \text{ moles of S}}{2 \text{ moles of Al}} \times 2.00 \text{ moles of Al} = 3.00 \text{ moles of S} \][/tex]

Since only 2.00 moles of sulfur are available (less than 3.00 moles), sulfur is the limiting reactant.

Step 3: Calculate the Moles of [tex]\(\text{Al}_2\text{S}_3\)[/tex] Produced

From the balanced equation, 3 moles of sulfur produce 1 mole of [tex]\(\text{Al}_2\text{S}_3\):[/tex]

[tex]\[ \text{Moles of } \text{Al}_2\text{S}_3 = \frac{2.00 \text{ moles of S}}{3} = 0.667 \text{ moles of } \text{Al}_2\text{S}_3 \][/tex]

Step 4: Convert Moles of [tex]\(\text{Al}_2\text{S}_3\)[/tex] to Grams

Calculate the molar mass of [tex]\(\text{Al}_2\text{S}_3\):[/tex]

  • Molar mass of [tex]\(\text{Al}\)[/tex] = 26.98 g/mol
  • Molar mass of [tex]\(\text{S}\)[/tex] = 32.07 g/mol

Molar mass of [tex]\(\text{Al}_2\text{S}_3\):[/tex]

  • [tex]\[ \text{Molar mass of } \text{Al}_2\text{S}_3 = 2 \times 26.98 + 3 \times 32.07 \][/tex]
  • [tex]\[ \text{Molar mass of } \text{Al}_2\text{S}_3 = 53.96 + 96.21 \][/tex]
  • [tex]\[ \text{Molar mass of } \text{Al}_2\text{S}_3 = 150.17 \text{ g/mol} \][/tex]

Calculate the mass of [tex]\(\text{Al}_2\text{S}_3\)[/tex] produced:

  • [tex]\[ \text{Mass} = 0.667 \text{ moles} \times 150.17 \text{ g/mol} \][/tex]
  • [tex]\[ \text{Mass} \approx 100.2 \text{ grams} \][/tex]

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Rewritten by : Barada

100.08 grams of aluminum sulfide can be produced when 2.00 mol of aluminum reacts with 2.00 mol of sulfur. Sulfur is the limiting reactant in this scenario.

To determine how many grams of aluminum sulfide (Al2S3) can be produced from 2.00 mol of aluminum (Al) reacting with 2.00 mol of sulfur (S), we first need to look at the balanced chemical equation:

2 Al + 3 S → Al2S3

This equation tells us that 2 moles of aluminum react with 3 moles of sulfur to produce 1 mole of aluminum sulfide. Given that we have only 2.00 mol of Al and 2.00 mol of S, sulfur is the limiting reactant because we need more sulfur to completely react with aluminum according to the stoichiometric ratio (2:3).

We first calculate the moles of Al2S3 produced:

2.00 mol S × (1 mol Al2S3 / 3 mol S) = 0.667 mol Al2S3

Molar mass of Al = 26.98 g/mol
Molar mass of S = 32.07 g/mol
Molar mass of Al2S3 = (2 × 26.98) + (3 × 32.07)

= 150.14 g/mol

Therefore, 0.667 mol of Al2S3 corresponds to:

0.667 mol × 150.14 g/mol = 100.08 g Al2S3

Thus, 100.08 grams of aluminum sulfide can be produced when 2.00 mol of aluminum reacts with 2.00 mol of sulfur.