4 Na + O 2 2 Na 2 O How many moles oxygen will react with 16.8 moles sodium? () 4 mol Na 1 mol O 2 16.8 mol Na = 4.20 mol O 2 O2O2 Na “Straight” Stoichiometry.

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4 Na + O 2 2 Na 2 O How many moles oxygen will react with 16.8 moles sodium? () 4 mol Na 1 mol O mol Na = 4.20 mol O 2 O2O2 Na “Straight” Stoichiometry -- the quantity of only one substance is given (no funny business) -- for gases, the conditions are STP

At STP, how many molecules of oxygen react with 632 dm 3 butane (C 4 H 10 )? C 4 H 10 O2O2 = 1.10 x m’c O 2 1 mol O 2 () 2 mol C 4 H mol O 2 () 1 mol C 4 H dm 3 C 4 H dm 3 C 4 H 10 __ C 4 H 10 + __ O 2 __ CO 2 + __ H 2 O x m’c O 2 () Suppose the question had been “how many ATOMS of O…” 1.10 x m’c O 2 () 1 m’c O 2 2 atoms O = 2.20 x at. O

At STP, how many molecules of oxygen react with 632 dm 3 butane (C 4 H 10 )? C 4 H 10 O2O2 = 1.10 x m’c O 2 1 mol O 2 () 2 mol C 4 H mol O 2 () 1 mol C 4 H dm 3 C 4 H dm 3 C 4 H 10 __ C 4 H 10 + __ O 2 __ CO 2 + __ H 2 O x m’c O 2 () Suppose the question had been “how many ATOMS of O…” 1.10 x m’c O 2 () 1 m’c O 2 2 atoms O = 2.20 x at. O

How many grams potassium will react with 465 grams nickel(II) phosphide? Ni 3 P 2 K 6 K + Ni 3 P 2 Ni + K 3 P 32 () = 458 g K 1 mol K 39.1 g K () 1 mol Ni 3 P 2 6 mol K () 1 mol Ni 3 P g Ni 3 P g Ni 3 P 2

limiting reactant (LR): the reactant that runs out first -- Any reactant you don’t run out of is an excess reactant (ER). Amount of EVERYTHING depends on the LR. Limiting Reactants (a.k.a., Limiting Reagents)

How to Find the Limiting Reactant For the generic reaction R A + R B P, assume that the amounts of R A and R B are given. Should you use R A or R B in your calculations? 1. Calc. # of mol of R A and R B you have. 2. Divide by the respective coefficients in balanced equation. 3. Reactant having the smaller result is the LR.

2 H 2 (g) + O 2 (g) 2 H 2 O(g) 13 g H g O 2 How many g H 2 O are formed? 2 1 mol H 2 2 g H 2 13 g H 2 () = 6.5 mol H 2 (HAVE) 1 mol O 2 32 g O 2 80 g O 2 () = 2.5 mol O 2 (HAVE). _.. _. 1= 2.50 = 3.25 LR = O 2 “Oh, bee- HAVE !” (And start every calc. with the LR.) = 90. g H 2 O () 1 mol O 2 1 mol H 2 O () 2.5 mol O 2 18 g H 2 O 2 mol H 2 O H2OH2OO2O2

2 Fe(s) + 3 Cl 2 (g) 2 FeCl 3 (s) 223 g Fe 179 L Cl 2 At STP, what is the limiting reactant? 2 1 mol Fe 55.8 g Fe 223 g Fe () = 4.0 mol Fe (HAVE) 1 mol Cl L Cl L Cl 2 () = 8.0 mol Cl 2 (HAVE). _.. _. 3= 2.66 = 2.0 LR = Fe What mass of FeCl 3 is produced? = 649 g FeCl 3 () 2 mol Fe 1 mol FeCl 3 () 4.0 mol Fe g FeCl 3 2 mol FeCl 3 FeCl 3 Fe (“Oh, bee-HAVE!”)