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Calculate the mass of excess reactant remaining after the reaction is comple

Question

Calculate the mass of excess reactant remaining after the reaction is comple

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Solution

Sure, I can help with that. To calculate the mass of the excess reactant remaining after the reaction is complete, follow these steps:

  1. Write the balanced chemical equation for the reaction.

  2. Determine the molar masses of the reactants involved.

  3. Calculate the moles of each reactant using their given masses and molar masses.

  4. Identify the limiting reactant by comparing the mole ratio of the reactants used in the reaction to the mole ratio in the balanced equation.

  5. Calculate the moles of the excess reactant that react using the stoichiometry of the balanced equation.

  6. Subtract the moles of the excess reactant that reacted from the initial moles of the excess reactant to find the moles of excess reactant remaining.

  7. Convert the remaining moles of the excess reactant back to mass using its molar mass.

Let's go through an example to illustrate these steps:

Example Problem:

Given the reaction: A+2B→C \text{A} + 2\text{B} \rightarrow \text{C}

Suppose you start with 10 grams of A and 15 grams of B. The molar masses are:

  • Molar mass of A = 50 g/mol
  • Molar mass of B = 30 g/mol

Step-by-Step Solution:

  1. Balanced Equation: A+2B→C \text{A} + 2\text{B} \rightarrow \text{C}

  2. Molar Masses:

    • A: 50 g/mol
    • B: 30 g/mol

  3. Calculate Moles:

    • Moles of A: 10 g50 g/mol=0.2 mol \frac{10 \text{ g}}{50 \text{ g/mol}} = 0.2 \text{ mol}
    • Moles of B: 15 g30 g/mol=0.5 mol \frac{15 \text{ g}}{30 \text{ g/mol}} = 0.5 \text{ mol}

  4. Identify Limiting Reactant: According to the balanced equation, 1 mole of A reacts with 2 moles of B.

    • Moles of B needed for 0.2 moles of A: 0.2 mol A×2 mol B/mol A=0.4 mol B 0.2 \text{ mol A} \times 2 \text{ mol B/mol A} = 0.4 \text{ mol B}
    • Since we have 0.5 moles of B, B is in excess and A is the limiting reactant.

  5. Calculate Moles of Excess Reactant that React:

    • Moles of B that react: 0.2 mol A×2 mol B/mol A=0.4 mol B 0.2 \text{ mol A} \times 2 \text{ mol B/mol A} = 0.4 \text{ mol B}

  6. Calculate Remaining Moles of Excess Reactant:

    • Initial moles of B: 0.5 mol
    • Moles of B that reacted: 0.4 mol
    • Remaining moles of B: 0.5 mol−0.4 mol=0.1 mol 0.5 \text{ mol} - 0.4 \text{ mol} = 0.1 \text{ mol}

  7. Convert Remaining Moles to Mass:

    • Mass of remaining B: 0.1 mol×30 g/mol=3 g 0.1 \text{ mol} \times 30 \text{ g/mol} = 3 \text{ g}

So, the mass of the excess reactant (B) remaining after the reaction is complete is 3 grams.

This problem has been solved

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