To determine how much H_2 O is in excess, use the 0.041 mol XeF_6 that you start

Question

To determine how much H_2 O is in excess, use the 0.041 mol XeF_6 that you start with, and the 1:3 mole relationship between XeF_6 and H_2 O in the balanced equation, to calculate the moles of H_2 O that will react. This number must be less than the original amount of 0.139 mol of H_2 O that you start with: 0.041 mol XeF_6 times 3 mol H_2 O/1 mol XeF_6 = 0.123 mol H_2 O reacts So of the original amount of 0.139 mol of H_2 O that you start with, only 0.123 reacts, and the rest is excess. Subtract to find the excess number of moles, and convert to grams to find the grams excess: 0.139 moles H_2 O start - 0.123 mole H_2 O reacts = 0.016 mol H_2 O excess 0.016 mol H_2 O excess times 18 g H_2 O/1 mol H_2 O = 0.288 g H_2 O excess 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

a)
moles of KCl reacted = mass of KCl / molar mass of KCl
= 55.1 / 74.55
= 0.739 mol
moles of KF formed = moles of KCl reacted
= 0.739 mol

b)
moles of CaF2 reacted = mass of CaF2 / molar mass of CaF2
= 46.7 / 78.07
= 0.598 mol
moles of KF formed = 2*moles of KCl reacted
=2* 0.598 mol
=1.196 mol

c)
KCl produces smaller product
so, KCl is limiting reagent

d)
moles of KCl reacted = 0.739 mol
moles of CaF2 reacted = 0.739/2 = 0.3695 mol
mass of CaF2 reacted = 0.3695*78.07 = 28.8 g

reamining mass of CaF2 = 46.7 - 28.8 = 17.9 g

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