Answer to Question #168728 in General Chemistry for Emily

Question

in the following reaction 7.5G of Cl2 and 4.6g of KBr are mixed in solution .calculate the number of moles of KBr and CI2. Determine the limiting excess reactants, how many grams would be left over of the excess reactant

Answer 1

Solution:

The balanced chemical equation:

2KBr + Cl₂ → 2KCl + Br₂

According to the equation: n(Cl₂) = n(KBr)/2

Determine moles of each reactant:

7.5 g Cl₂ × (1 mol Cl₂ / 70.906 g Cl₂) = 0.1058 mol Cl₂

Moles of Cl₂ = 0.1058 mol

4.6 g KBr × (1 mol KBr / 119.002 g KBr) = 0.03865 mol KBr

Moles of KBr = 0.03865 mol

Choose one reactant and determine how many moles of the other reactant are necessary to completely react with it. Let's choose KBr:

n(Cl₂) = n(KBr)/2 = 0.03865 mol / 2 = 0.019325 mol

The calculation above means that we need 0.019325 mol of Cl₂ to completely react with KBr.

We have 0.1058 mol Cl₂ and therefore more than enough chlorine.

Thus chlorine (Cl₂) is in excess and potassium bromide (KBr) must be the limiting reactant.

Hence,

KBr - limiting reactant

Cl₂ - excess reactant

Finally, we can find how many grams would be left over of the excess reactant:

0.1058 mol Cl₂ to start - 0.019325 mol of Cl₂ needed = 0.086475 mol Cl₂ in excess

0.086475 mol Cl₂ × (70.906 g Cl₂ / 1 mol Cl₂) = 6.13 g

m_excess(Cl₂) = 6.13 g

Answer:

Moles of Cl₂ = 0.1058 mol;

Moles of KBr = 0.03865 mol;

Cl₂ - excess reactant;

KBr - limiting reactant;

6.13 g Cl₂ would be left over.