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The reaction of boron and chlorine can be expressed as the following chemical equation:

2 B + 3 Cl₂ ==> 2 BCl₃
If 44.2 grams of boron completely reacts with 178.2 grams of chlorine gas, then how many grams of the EXCESS REACTANT will be leftover?

1 Answer

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Final answer:

To calculate the excess reactant leftover after a reaction between boron and chlorine, write the balanced equation, determine the limiting reagent using stoichiometry, and subtract the amount of chlorine that reacted from the initial amount to find the leftover mass.

Step-by-step explanation:

Calculating the amount of excess reactant left over after a chemical reaction requires understanding of stoichiometry, limiting reactants, and theoretical yield.

First, we must write the balanced equation for the reaction between boron (B) and chlorine (Cl₂).

B + 3Cl₂ → 2BCl₃

Using the molar masses of boron (10.81 g/mol) and chlorine (70.90 g/mol), calculate the moles of each reactant:

  • Moles of boron = 44.2 g / 10.81 g/mol
  • Moles of chlorine = 178.2 g / 70.90 g/mol

Next, determine the limiting reagent by comparing the stoichiometry of the balanced equation. For every mole of boron, 3 moles of chlorine are required. Calculate the moles needed of chlorine for the moles of boron available:

  • Moles of Cl₂ required = Moles of B x 3

If the calculated moles of chlorine required are greater than the moles of chlorine available, boron is the limiting reagent. Otherwise, chlorine is limiting.

After identifying the limiting reagent, calculate how much of the non-limiting reagent (the excess reagent) remains by subtracting the amount that reacts from the total amount present.

Finally, convert the remaining moles of the excess reagent back to grams using its molar mass. This will give you the mass of excess reactant leftover after the reaction.

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