 # Chapter 3 Calculations with Equations & Concentrations.

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Chapter 3 Calculations with Equations & Concentrations

Calculations with Equations Butane, C 4 H 10, is completely combusted with excess oxygen. If 17.0 g of butane react, how much O 2 is required? Ans: 61 g

Calculations with Equations How much lithium chloride is produced when 100 g each of lithium and table salt react? Ans: 72.3 g

Calculations with Equations What is the limiting reactant when 20 g each of manganese II oxide, potassium hydroxide oxygen and chlorine react to produce potassium permanganate, potassium chloride and water? Ans: KOH is limiting

Calculations with Equations If 84 g of bromine gas reacts w/ excess chlorine gas to form bromine chloride, what is the percent yield if 107.0 g of bromine chloride is produced? Ans: 88.4%

Calculations with Sequential Equations How many grams of ammonia, NH 3, is required to produce 1 mole of ammonium nitrate, NH 4 NO 3 by the following reaction pathway? 4NH 3 + 5O 2  4NO + 6H 2 O 2NO + O 2  2NO 2 3NO 2 + H 2 O  2HNO 3 + NO HNO 3 + NH 3  NH 4 NO 3 1 mol

Calculations with Sequential Equations HINTs: Start at the end and work backwards! Work in moles first. Convert to grams for your final answer. Ans: 42.5 g

Concentration of Solutions Many rxn. Use water or some other liquid as a reaction medium. The reaction medium allows the reagents to get close enough to each other to chemically react. Solution – homogeneous mixture composed of a dissolving medium [solvent] and, the substance which is dissolved [solute].

Aqueous solution – a solution in which water is the solvent. The amount of solute dissolved in a given amount of solvent is referred to as the solutions concentration. The more solute the greater the concentration.

Expressing Concentration There are several methods of expressing the concentration of a solution. Molarity (M) –M = moles solute – liter solution Molality (m) –m= moles solute – kg solvent Percent Mass (%) – % = mass solute 100 = _______mass solute_____ 100 mass solution mass solute + mass solvent Normality (N) – N = equivalents solute = equivalents of solute___ liter solutionliters solute + liters solvent

Expressing Concentration equivalent – the mass of an acid or base that produces one mole [6.02 x 1023] H+ or OH- ions when placed in solution. HCl NaOH 1 equivalent acid 1 equivalent base H 2 SO 4 Ca(OH) 2 2 equivalent acid 2 equivalent base H 3 PO 4 Al(OH) 3 3 equivalent acid 3 equivalent base

Determine the molarity of a solution containing 500g of phosphoric acid dissolved in 3.00 L of solution. Molarity (M) M = moles solute liter solution 500 g H 3 PO 4 │ 1mol H 3 PO 4 = 5.10 mol H 3 PO 4 │ 98 g M = 5.10 mol = 1.7 M H3PO4 3 L

Determine the molality of a solution containing 500g of phosphoric acid dissolved in 3.00 L of solution. Molality (m) m= moles solute kg solvent m = 5.01 mol______ = 2.04 m H 3 PO 4 3 kg soln -.5 kg solute *If the solvent is not specified assume that it is water.

Determine the % mass of a solution containing 500g of phosphoric acid dissolved in 3.00 L of solution. Percent Mass (%) % = mass solute 100 = _ mass solute_____ 100 mass solution mass solute + mass solvent % = 500 g H3PO4 100 = 16.7% H3PO4 3000 g soln

Determine the normality of a solution containing 500g of phosphoric acid dissolved in 3.00 L of solution. Normality (N) N = equivalents solute = equivalents of solute___ liter solutionliters solute + liters solvent 5.10 mol H 3 PO 4 │ 3 equiv. H+ = 15.3 equiv. │ mol H 3 PO 4 15.3 equiv H 3 PO 4 = 5.1 N H3PO4 3.00 L soln

How much 10.0 M NaHClO2 [sodium hypochlorite; a.k.a. bleach] is required to prepare 375 ml of a.60 M solution of bleach? M = moles solute liter solution (M)(liter solution) = moles solute If additional solvent (L) is added to the 10.0 M bleach solution, the concentration of the bleach (M) will decrease, but the amount of [atoms] bleach(mol) will remain constant. Thus… (M)1(liter solution)1 = moles solute = (M)2(liter solution)2 Or (M) 1 (liter solution) 1 = (M) 2 (liter solution) 2

This formula can be used to solve molarity dilution calculations. For the calculation in question… (M)1(liter solution)1 = (M)2(liter solution)2 (X) (10.0 L) = (.60 M)(.375 L) X =.0225 L = 22.5 mL

Stock solutions of HNO3 are 69.6% nitric acid by mass. What is the molarity of the solution if the specific gravity of the solution is 1.45? S.G. = Density standard for solid / liquid = 1.0 g/mL Standard standard for gases = 1.29 g/L 1.45 = Density 1.0 g/mL D = 1.45 g/mL M = moles solute liter solution Converting g/mL to mol/L… 1.45g │ 1 mol │ 1000 mL =23.02 M, if 100% nitric acid mL │ 63 g │ 1 L (23.02 mol/ L)(.696) = 16 M

Given the reaction H 2 CO 3 \ Na 2 CO 3 + 2HCl  2NaCl + CO 2 + H 2 O How much.75 M acid is required to react with 4.25 g of sodium carbonate? Na2CO3 + 2HCl  2NaCl + CO2 + H2O.4.25 g x 4.25 g Na2CO3 │ 1 mol │ 2 mol HCl = 8.02 x 10-2 mol HCl │ 106 g │ 1 mol Na2CO3 M = moles solute liter solution.75 M =.75 mol = 8.02 x 10-2 mol HCl 1 L soln x =.107 L acid soln