How much magnesium sulphide can be obtained from 2g of magnesium and 2g of sulphur by the reaction Mg+S ––> MgS?What is the limiting reactant?Calculate the amount of one of the reactants which remains unreacted.

To determine how much magnesium sulfide (MgS) can be produced from 2 grams of magnesium (Mg) and 2 grams of sulfur (S), we first need to look at the balanced chemical equation for the reaction:

Balanced Reaction

The reaction between magnesium and sulfur can be represented as:

Mg + S → MgS

This equation shows that one mole of magnesium reacts with one mole of sulfur to produce one mole of magnesium sulfide.

Molar Mass Calculation

Next, we need to calculate the molar masses of magnesium and sulfur:

• Molar mass of magnesium (Mg): approximately 24.31 g/mol
• Molar mass of sulfur (S): approximately 32.07 g/mol

Convert Grams to Moles

Now, we can convert the given masses of magnesium and sulfur into moles:

• Moles of magnesium = mass (g) / molar mass (g/mol) = 2 g / 24.31 g/mol ≈ 0.0822 moles
• Moles of sulfur = mass (g) / molar mass (g/mol) = 2 g / 32.07 g/mol ≈ 0.0623 moles

Identifying the Limiting Reactant

To find the limiting reactant, we compare the mole ratio of the reactants to the stoichiometry of the balanced equation. According to the equation, 1 mole of magnesium reacts with 1 mole of sulfur. We have:

• 0.0822 moles of magnesium
• 0.0623 moles of sulfur

Since we need equal moles of both reactants, sulfur is the limiting reactant because we have fewer moles of it compared to magnesium.

Calculating the Amount of Magnesium Sulfide Produced

Since sulfur is the limiting reactant, the amount of magnesium sulfide produced will be based on the moles of sulfur available:

• From the reaction, 1 mole of S produces 1 mole of MgS.
• Thus, 0.0623 moles of sulfur will produce 0.0623 moles of magnesium sulfide.

Now, we can convert moles of magnesium sulfide back to grams using its molar mass:

• Molar mass of magnesium sulfide (MgS) = 24.31 g/mol (Mg) + 32.07 g/mol (S) = 56.38 g/mol
• Mass of MgS produced = moles × molar mass = 0.0623 moles × 56.38 g/mol ≈ 3.52 g

Amount of Unreacted Magnesium

To find out how much magnesium remains unreacted, we need to determine how much magnesium was consumed in the reaction:

• Since 0.0623 moles of sulfur reacted, it will consume 0.0623 moles of magnesium (1:1 ratio).
• Initial moles of magnesium = 0.0822 moles
• Moles of magnesium remaining = initial moles - moles reacted = 0.0822 moles - 0.0623 moles = 0.0199 moles

Now, converting the remaining moles of magnesium back to grams:

• Mass of unreacted magnesium = moles × molar mass = 0.0199 moles × 24.31 g/mol ≈ 0.48 g

Summary of Results

In summary, from the reaction of 2 grams of magnesium and 2 grams of sulfur:

• Amount of magnesium sulfide produced: approximately 3.52 grams
• Limiting reactant: sulfur
• Unreacted magnesium remaining: approximately 0.48 grams

This analysis illustrates how to approach stoichiometric calculations in chemical reactions by identifying limiting reactants and calculating the amounts of products and remaining reactants. If you have any further questions or need clarification, feel free to ask!