Stoichiometry help please

Question

Given: 2K + Cl2 = 2KCl

1. If 0.143 grams of K and 0.236 grams of Cl2 react to form Potassium Chloride, how much Potassium Chloride can be formed? (Hint - you are finding the limiting reactant.)
2. How much excess reactant is left over?
3. If 0.200 grams of KCl are actually formed in lab, what is the percent yield? (%yield - actual yield / theoretical yield *100)

Answer 1

We will need moles of both K and Cl₂ since the chemical reaction equation uses moles not grams.

To convert grams to moles, we divide mass by the molar mass. The molar mass of potassium (K) from the Periodic Table is 39.098 g/mol so 0.143 grams of potassium is 0.003657 moles. The molar mass of Chlorine from the Periodic Table is 35.453 g/mol and since there are 2 atoms in Cl₂, the molar mass of Cl₂ is 70.906 g/mol so 0.236 g of Cl₂ is 0.003328 moles.

The chemical reaction equation says I need 2 moles of K to react with 1 mole of Cl₂ and since there are approximately equal amounts (moles) of both K and Cl2, obviously the potassium is the limiting reactant (there is excess Cl₂ when I use all the K up).

That means the roadmap I will use to determine the amount of KCl produced looks like this:

We've already calculated that 0.143 g of K is 0.003657 moles. Since 2 moles of K are needed to 1 mole of Cl₂ then 0.003657/2 or 0.0018288 moles of Cl₂ was needed in the reaction. Since there was 0.003328 moles of Cl₂ present then 0.003328 - 0.0018288 or 0.0014996 moles did not get used in the reaction. multiplying by the molar mass of 70.906 g/mol means the there was an excess of 0.106 g of Cl₂ left over.

For question b), the moles of KCl produced equals the moles of K reacted, so 0.003657 moles. The molar mass of KCl is 39.098 + 35.453 = 74.551 g/mol so the mass of KCl that should have been produced was (0.003657 moles)(74.551 g/mol) = 0.27267 g of KCl. That make the percent yield = 0.200/0.27267( x 100) = 73.3%