Unit 10: Stoichiometry Using mole ratios and review of balancing chemical equations

Stoichiometry: The study of quantities as it relates to chemical reactions. The word stoichiomety comes from the Greek words: stoicheion (meaning "element") metron (meaning "measure")

Balanced chemical equations can be interpreted many ways… 1N 2(g) + 3H 2(g)  2NH 3(g) molecules: NN HH +  HHHH N HH H N HH H

Balanced chemical equations can be interpreted many ways… mass: +  28.02g 6.06g 34.08g = Mass of reactants and products are always equal! 1N 2(g) + 3H 2(g)  2NH 3(g)

Balanced chemical equations can be interpreted many ways… moles: +  1 mol N 2 3 mol H 2 2 mol NH 3 This is the relationship we will focus on in this unit! 1N 2(g) + 3H 2(g)  2NH 3(g)

Mole-Mole Calculations The coefficients in the balanced equation represent the smallest whole number mole ratio between reactants and products.

Mole-Mole Calculations Examples of mole ratios for the reaction: 1N 2(g) + 3H 2(g)  2NH 3(g) 1 mol N 2 2 mol NH 3 3 mol H 2 1 mol N 2 2 mol NH 3 2 mol H 2

Mole-Mole Calculations Example: If 3.86 moles of potassium reacts completely with excess water, how many moles of hydrogen would be produced? potassium water  + potassium hydroxide + hydrogen hydrogen hydroxide __K + __ H(OH)  __ K(OH) + __ H K: 3.86 mol K U: ? mol H mol K 1 x 1 mol H 2 2 mol K = 1.93 mol H 2

Mole-Mole Calculations Example: How many moles of aluminum will react with moles of hydrochloric acid? aluminumhydrochloric acid  + aluminum + hydrogen chloride hydrogen chloride __Al + __HCl  __ AlCl 3 + __ H K: mol HCl U: ? mol Al mol HCl 1 x 2 mol Al 6 mol HCl = mol Al

Unit 10: Stoichiometry Mass-mass calculations using the mole ratio

There are three steps needed for these calculations: 1.Convert grams  moles (Using the MM of the known.) 2.Convert moles of known  moles of unknown using the mole ratio (We just learned this.) 3.Convert moles of unknown  grams (Using MM of the unknown.)

Mass-Mass Calculations Example: What mass of silver chloride will react with 15.0g of aluminum? __Al + __ AgCl  __ Ag + __ AlCl K: 15.0 gAl U: ? gAgCl 15.0 gAl 1 x 3 mol AgCl 1 mol Al = 239g AgCl 1 mol Al gAl x x gAgCl 1 mol AgCl 1Ag = ) 1Cl =35.45(1) g Al Ag Cl Review!

Mass-Mass Calculations Example: What mass of oxygen gas would be produced from the decomposition of 37.2g of lithium chlorate? __Li(ClO 3 )  __ LiCl + __ O K: 37.2 gLi(ClO 3 ) U: ? gO gLi(ClO 3 ) 1 x 3 mol O 2 2 mol Li(ClO 3 ) = 19.8 gO 2 1 mol Li(ClO 3 ) gLi(ClO 3 ) x x gO 2 1 mol O 2 1Li =6.94(1) 1Cl =35.45(1) 3O =16.00(3) 90.39g

SUMMARY OF IMPORTANT POINTS: You are calculating the mass for TWO different substances! Molar mass (MM) is ALWAYS equal to ONE mole! The ONLY time the number of moles is different than ONE (1) is when you are doing the MOLE RATIO STEP!