The Math of Equations Stoichiometry the calculation of quantities in chemical reactions

Limiting Reactant Calculate the amount of product that could be made for each reactant Compare the amount that could be made to the amount of product actually made to determine which reactant is limiting Calculate the excess reactant by calculating the amount that actually reacts and then subtracting

Limiting Reactant The Chemical used up first is the LIMITING REACTANT There is always some unreacted chemical Because you cannot measure to the last atom So one reactant gets used up first Once one reactant is used up, no more product can be made The Chemical used up first is the LIMITING REACTANT

Excess reactant So…….. reactant that is not all used up in the reaction

2H2 +O2 2H2O 1. Determine the limiting reactant when.. 4 grams of H2 are reacted with 31g of O2. Determine moles of products each reactant can form The one that is limiting will form the least amount of product 4g H2 x 1 mole H2 x 2 moles H2O = 2 moles H2O 2 g H2 2 moles H2 31g O2 x 1 mole O2 x 2 moles H2O = 1.93 moles H2O 32 g O2 1 moles O2 Oxygen would run out first, that’s why it makes less water. Oxygen is the limiting reactant

Mg + 2HCl  MgCl2 + H2 3.500 g of Mg is added to 300.0 mL of 1.000 M HCl How many grams of Magnesium chloride are formed? 3.500g Mg x 1 mole x 1 mole MgCl2 x 95.21 g MgCl2 = 13.71g MgCl2 24.31 1 mole Mg mole MgCl2 0.3000L x 1.000mole x 1 mole MgCl2 x 95.21 g MgCl2 = 14.28gMgCl2 1 L 2 mole HCl mole MgCl2 Since Mg “runs out first”, it is the limiting reactant. What test could you run to determine there was acid left over?

Mg + 2HCl  MgCl2 + H2 3.5 g of Mg is added to 300.0 mL of 1.0M HCl How many liters of gas @ STP? Since magnesium is the limiting reactant 3.500g Mg x 1 mole x 1 mole H2 x 22.4 L H2 = 6.450 L H2 24.31 1 mole Mg mole H2

Mg + 2HCl  MgCl2 + H2 What is left over and how many moles? If magnesium is the LR, then HCl is left over 3.500g Mg x 1 mole x 2 mole HCl = .2879 moles HCl used in reaction 24.31 1 mole Mg 0.3000L x 1.000mole HCl = .3000 moles HCl given 1 L 0.3000 moles HCl - 0.2879 (moles HCl used by Mg) = .0121 mols HCl 0.0121moles HCl = .0403 M .0121moles HCl x 36.5g = 0.4417g 0.3000L mole HCl

Interpreting Balanced equations 2H2O  2H2 + O2 2 molecules  2 molecules + 1 molecule 2 mole  2 mole + 1 mole

Steps in stoichiometry Write symbols and balance Write the moles on top/known on bottom Circle the known and unknown Do Dimensional analysis problem

Mole – Mole problems Ammonium nitrate decomposes to yield dinitrogen monoxide and water How many moles of water are produced if 14.8 moles of ammonium nitrate react? Write symbols and balance Write the moles on top/known on bottom Circle the known and unknown Do Dimensional analysis problem

practice Bubbles of hydrogen gas and iron(III) chloride are produced when iron is dropped into hydrochloric acid How many moles of hydrogen are produced when 14.5 moles of iron react?

Mole – Mass problems Aluminum and silver nitrate react. How many moles of aluminum nitrate will be produced if 34.6 g of aluminum react? Write symbols and balance Write the moles on top/known on bottom Circle the known and unknown Do dimensional analysis problem

Mass-Mass Lithium nitride reacts with water to produce ammonia and lithium hydroxide. How many g of LiOH are produced if .38g of Li3N react?

Other Sodium chloride is produced when chlorine reacts with sodium 1.How many L of Chlorine are needed to make 45.6 g sodium chloride? 2.How many F.U. of NaCl are produced when 1.32 x 1015 atoms of Na react?

Law of Conservation of Matter Prove the law of conservation of matter for 2NH3  N2 + 3H2 Do Dimensional analysis problem for each reactant and product to determine the mass in g

% Yield (product) % Yield = Actual (experiment) x 100% (product) Theoretical (calculation)

You Try 13.5g Na x 1 mol Na x 1 mol H2 x 22.4L H2 = 13.5 g of sodium metal reacts with excess hydrochloric acid. How many liters of hydrogen are produced (at STP… 22.4L / mol) 2Na + 2HCl  2 NaCl + H2 13.5g Na x 1 mol Na x 1 mol H2 x 22.4L H2 = 23.0 g Na 2 mol Na 1 mol H2 6.57 liters of H2 gas Steps to solve Predict products Balance equation Set up Dimensional analysis Solve

4Al + 3O2  2Al2O3 15.0g x 1 mole 27.01 g x 2 mol Al2O3 4 mole Al 15.0 g of aluminum are reacted with excess oxygen. How much oxide was produced? 4Al + 3O2  2Al2O3 15.0g x 1 mole 27.01 g x 2 mol Al2O3 4 mole Al x 102.g Al2O3 = 1 mol Al2O3 28.3 g aluminum oxide

Four Moles of H2O! 2H2 + O2 —> 2H2O Matter is Conserved 4g + 32g = 36g 8g + 64g = ??g 8g + 64g = 72g If you react 8 grams of hydrogen and 64 g of oxygen, how much water would did you make? Four Moles of H2O! Or 4 x 18 g/mole = 72 grams

2H2 + O2 —> 2H2O 2 moles of H2 = how many grams? How many grams of water are produced from 2 moles of hydrogen gas and 1 mole of oxygen gas? 2H2 + O2 —> 2H2O 4g 32g 36g 2 moles of H2 = how many grams? 2 mole x 2 atoms x 1 g/mole = 4g 1 mole of O2 = how many grams? 2 atom x 16 g/mole = 32 g 2 mole of H2O = how many grams 2 mole x 18 g/mole = 36g