For the equation 2 KCl + CaF_2 rightarrow 2 KF +Cacl_2, 55.1 grams of KCl is rea
ID: 505053 • Letter: F
Question
For the equation 2 KCl + CaF_2 rightarrow 2 KF +Cacl_2, 55.1 grams of KCl is reacted with 46.7 grams CaF_2. a. How many moles of KF will be produced if all of the KCl reacts? b. How many moles of KF will be produced if all of the CaF_2 reacts? c. Which reactant, KCl or CaF_2, produces the smaller amount of product? This reactant is called the limiting reactant or limiting reagent, and controls how much of the other reagent is consumed. d. Based on this limiting reagent, how much of the other reagent is actually consumed and how much is in excess?Explanation / Answer
2 KCl + CaF2 -----> 2 KF + CaCl2
2 moles 1 mole 2 mole 1 mole
Given
Mass of KCl = 55.1 g
mass of CaF2 = 46.7 g
Molar mass of KCl = 74.55 g/mol
Molar mass of CaF2 = 78 g/mol
No. of moles of KCl = Mass of KCl / Molar mass of KCl = 55.1 g / 74.55 g/mol = 0.739 moles
No. of moles of CaF2 = Mass of CaF2 / Molar mass of CaF2 = 46.7 g / 78 g/mol = 0.599 moles
according to reaction stoichometry when 2 moles of KCl reacts 2 moles of KF will be formed so
when 0.739 moles of KCl reacts it will form 0.739 moles of KF Answer (a)
according to reaction stoichometry when 1 mole of CaF2 reacts it will form 2 moles of KF
when 0.599 moles of CaF2 reacts it will form 2*0.599 = 1.198 moles of KF Answer (b)
KCl produces small amount of CaF2 so this the limiting reactant Answer (c)
for every 2 moles of KCl 1 mole of CaF2 is consumed according to reaction stoichometry
we have 0.739 moles of KCl reacting so
No. of moles of CaF2 consumed = No. of moles of KCl / 2 = 0.739 /2 = 0.3695 moles of CaF2 Answer (d)
excess = 0.599 - 0.3695 = 0.2295 moles of CaF2 Answer (d)