Limiting Reactants Scenario #1: Reactants are in 1:1 Ratio –Reactant with smallest no. moles is limiting. A + B →C 0.5 moles 0.25 moles? moles –The answer has to be 0.25 moles because when reactant B is used up, the reaction STOPS!
Limiting Reactants Scenario #2: Reactants are NOT in 1:1 Ratio –Must find number of moles product using each reactant quantity and their coefficient in balanced reaction to establish which reactant is limiting. A + 3 B →2 C 2.50 moles 4.50 moles? moles –How many moles product will form? Ans. = 3.00 moles.
Limiting Reactants Scenario #2: Reactants are NOT in 1:1 Ratio A + 3 B →2 C 2.50 moles 4.50 moles? moles –2.50 moles A x (2 moles C / 1 moles A) = 5.00 mol C –4.50 moles B x (2 moles C / 3 moles B) = 3.00 mol C –B is limiting reactant; it forms least amount product. –B reacts three times for every A that reacts; this occurs despite having more moles when compared to reactant A.
Example Problem If we react 42.5 g Mg and 33.8 g O 2, what is limiting reactant & theoretical yield? Reaction:2 Mg + O 2 ==> 2 MgO
Solutions: An Introduction A solution is a homogeneous mixture that consists of two components. –Solute is substance being dissolved in the solution. –Solvent is the dissolving environment. –Most chemical reactions are carried out in liquid state or in solution. –In biology & general chemistry, your solvent is water unless told otherwise.
Solutions: An Introduction Molarity is an expression of solution concentration. It is defined as the number of moles of solute divided by the volume of a solution in liters. –Molarity = M = moles solute / Vol. soln. in Liters –By simple rearrangement one obtains the following equations. –(Vol. soln. in Liters ) x (M) = moles solute –Vol. soln. in Liters = moles solute / M
Solutions: An Introduction A dilution is a process by which water (or another solvent) is added to a solution of known concentration to achieve a new solution of lower concentration. –Dilution does not alter number of moles of solute. –Dilution equation M 1 V 1 = M 2 V 2
Solutions: An Introduction Some example problems to consider. What is the molarity of a solution prepared by adding 24.66 g of potassium nitrate to 250 mL of water? –Answer = 0.975 M How does one prepare 250 mL of 1.50 M HCl from 12.0 M HCl ? –Answer = Measure 31.25 mL of 12.0 M HCl and add to 218.75 mL of distilled water. Mix contents well before using.
Solution Stoichiometry Handy Dandy Five Step Method is still used. –The only modification is now we have volumes and concentrations to consider. –Use Molarity equations to find number of moles and then proceed as before.
An example involving solutions. –Sodium hydrogen carbonate (NaHCO 3 ) is used as an antacid. One tablet requires 34.50 mL of 0.138M HCl solution for complete reaction. Determine the number of grams of NaHCO 3 that one tablet contains: –The balanced chemical reaction is shown below. NaHCO 3 + HCl → NaCl + H 2 O + CO 2 Answer:0.400 grams NaHCO 3