Stoichiometry

Limiting Reagent What happens in a chemical reaction, if there is an insufficient amount of one reactant?

In the laboratory, a reaction is rarely carried out with exactly the required amounts of each of the reactants. In most cases, one or more reactants is present in excess. Once one of the reactants is used up, no more product can be formed.

Limiting Reagent: the reagent that is completely used up in a chemical reaction. Excess Reagent: reagent not completely used up in a chemical reaction.

Example: 4 NH3 + 5 O2  4 NO + 6 H2O If 750g of each reagent is mixed, how much NO can be theoretically formed? What is the limiting reagent and what is the excess reagent. How much NH3 is actually used and how much is left over?

4 Na (s) + O2(g)  2 Na2O (s) 23 g Na 4 mole Na 1 mol Na2O Sodium metal reacts with oxygen to produce sodium oxide. If 5.00 g of sodium reacted with 5.00 grams of oxygen, how many grams of product is formed? 4 Na (s) + O2(g)  2 Na2O (s) 5.00g Na (1 mole Na) ( 2 mole Na2O)( 62 g Na2O) = 6.74 g of Na2O 23 g Na 4 mole Na 1 mol Na2O 5.00g O2 (1 mole O2) ( 2 mole Na2O)( 62 g Na2O) = 19.38 g of Na2O 32 g O2 1 mole O2 1 mol Na2O Na (sodium) “limits” how much sodium oxide is produced. The correct answer is 6.74 g of sodium oxide.

4 Na (s) + O2(g)  2 Na2O (s) 23 g Na 4 mole Na 1 mol O2 How much oxygen was used in this reaction and how much of each reactant was leftover (in excess)? 4 Na (s) + O2(g)  2 Na2O (s) Use stoichiometry to compare the two reactants. The amount of O2 used to make 6.74 g of Na2O is calculated by: 5.00g Na (1 mole Na) ( 1 mole O2)( 32 g O2) = 1.74 g of O2 was used 23 g Na 4 mole Na 1 mol O2 The amount of oxygen (O2) leftover can be calculated by subtracting the starting mass of oxygen from the used mass. 5.00g – 1.74 g = 3.26 g in excess The amount of sodium (Na) leftover at the end of the reaction is “0.00 g” (zero), since it was the limiting reactant and was completely consumed in the reaction.

Example 4.3, page 168 Ni(II)S occurs naturally as part of the mineral millerite. To analyze a mineral sample for the quantity of Ni(II)S, the following reactions are performed. NiS + 4 HNO3 Ni(NO3)2 + S + 2 NO2 + 2 H2O Ni(NO3)2 + 2 C4H8N2O2 Ni(C4H7N2O2)2 + 2 HNO3 Suppose a 0.468 g sample of the mineral millerite produces 0.206 g of Ni(C4H7N2O2)2. What is the mass percent of Ni(II)S in the mineral?

Example, page 169 Suppose a 2.367 g of an impure TiO2 sample evolves 0.143 g of O2 gas. What is the mass percent of TiO2 in this impure sample. 3 TiO2 + 4 BF3 3 TiF4 + 2 Br2 + 3 O2

Preparing Solutions

Empirical Formula by Combustion Analysis Example 4.4 (page 170) A hydrocarbon CxHy, was burned in oxygen and produces 3.447g of CO2 and 2.647g of H2O. Determine the empirical formula. If the molar mass of the compound is 86.2 g/mol, determine the molecular formula.

Empirical Formula by Combustion Analysis Example 4.5 (page 171) An isolated compound contains only C, H and O. Heating 0.513g of the compound gives: 0.501g of CO2 and 0.103g of H2O. Determine the empirical formula.

Concentration (M) Absorbance A solution of KMnO4 has an absorbance of 0.539 when measured at 540 nM in a 1.0 cm cell. What is the concentration of the KMnO4? The following data has been collected: Concentration (M) Absorbance 0.0300 0.162 0.0600 0.330 0.0900 0.499 0.120 0.670 0.150 0.840

Slope = 5.653 y intercept = -0.009

y = mx + b y – absorbance of unknown sample m – slope of line b – y intercept x – concentration of unknown sample 0.539 = 5.653 (x) + (-0.009) x = 0.0969 M