1. K + Cl2 —> KCI
Given: 180 g K and 120 g CI2

A.  Cl₂

B.  51.36 g K

C.  245.3 g KCl

Explanation:

Balance the given chemical equation

K  +  Cl₂  ⇒  KCl

2 K  +  Cl₂  ⇒  2 KCl

A. To find the limiting reagent, first convert the mass of each reactant to moles.  The molar masses are 39.10 g/mol for K and 70.91 g/mol for Cl₂.

(180 g)/(39.10 g/mol) = 4.60 mol K

(120 g)/(70.91 g/mol) = 1.69 mol Cl₂

Now, convert moles of each reactant to moles of the product.  You can do this by using the mole ratio in the balanced chemical reaction.  The reactant with the smallest moles of product is the limiting reagent.

(4.60 mol K) × (2 mol KCl/2 mol K) = 4.60 mol KCl

(1.69 mol Cl₂) × (2 mol KCl/1 mol Cl₂) = 3.29 mol KCl

Cl₂ is the limiting reagent.

B.  You need to find out how much excess K you will have.  Since Cl₂ is the limiting reagent, you will only make 3.29 mol of KCl.  Convert this value to moles of K using the mole ratio.

(3.29 mol KCl) × (2 mol K/2 mol KCl) = 3.29 mol K

Now convert moles of K to grams.  The molar mass is 39.10 g/mol.

(3.29 mol K) × (39.10 g/mol) = 128.64 g K

This is how much K will be consumed in the reaction.  Subtract the total amount of K by the amount that will be consumed.

180 - 128.64 = 51.36 g

There will be 51.36 g of excess K.

C.  You know that you will produce 3.29 mol of KCl based on previous calculations.  Convert this to grams.  The molar mass is 74.55 g/mol.

(3.29 mol) × (74.55 g/mol) = 245.3 g KCl

You will produce 245.3 g of KCl.

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