Chapter 9 - Stichimetry Sectin 9.1 Intrductin t Stichimetry Types f Stichimetry Prblems Given is in mles and unknwn is in mles. Given is in mles and unknwn is in mass (grams). Given is in mass and unknwn is in mles. Given is in mass and unknwn is in mass. Writing and Using Mle Ratis Mle Rati: a cnversin factr derived frm the cefficients f a balanced chemical equatin interpreted in terms f mles. In chemical calculatins, mle ratis are used t cnvert between: N 2 (g) + 3H 2 (g) 2NH 3 (g) Three mle ratis can be derived frm the balanced equatin abve: Example: Li(s) + O 2 (g) Li 2 O(s) Sectin 9.2 Ideal Stichimetric Calculatins Mle-Mle Calculatins Hw many mles f ammnia (NH 3 ) are prduced when 0.60 ml f N 2 reacts with H 2. Step 1: Write ut equatin and balance it Step 2: Determine mle rati f knwn (N 2 ) t unknwn (NH 3 ). Step 3: Multiply given amunt f knwn by the crrect mle rati that will cancel ut the knwn units, leaving yu the unknwn units 1
Example: Hw many mles f Chlrine are prduced when 3.2 mles sdium chlride decmpses? Hmewrk: pg 301 #2 and pg 311 #1 a, b (balance reactin and then a & b each have 3 answers) Mle-Mass Calculatins Hw many grams f aluminum xide(al 2 O 3 ) are prduced when 0.50ml Al reacts with O 2? Step 1: Write ut equatin and balance it Step 2: Multiply by the mle rati t get frm knwn t unknwn mles. Step 3: Multiply the mles f unknwn by the mlar mass f the unknwn t get grams. Example: Hw many grams f Cpper are needed t react with Sulfur t frm 5.40 mles f cpper sulfide? Mass-Mle Calculatins Calculate the number f mles f H 2 O prduced by the reactin f 5.40 g f O 2 with an excess f H 2. *excess means that there is enugh t react. Step 1: Write ut equatin and balance it. Step 2: Multiply the given amunt by the mlar mass f the knwn t get mles. Step 3: Multiply by the mle rati t get frm knwn t unknwn mles. Example: Hw many mles f hydrgen are needed t react with nitrgen t frm 10.5 grams f ammnia (NH 3 )? 2
Example: Hw many mles f Lithium are frmed by the decmpsitin f 3.40 grams f Lithium Oxide? Mass-Mass Calculatins Calculate the number f grams f CO prduced by the reactin f 3.25 g f C with an excess f O 2. Step 1: Write ut equatin and balance it. Step 2: Multiply the given amunt by the mlar mass f the knwn t get mles. Step 3: Multiply by the mle rati t get frm knwn t unknwn mles. Step 4: Multiply the mles f unknwn by the mlar mass f the unknwn t get grams. Example: Hw many grams f Oxygen are needed t make 20.0 grams f NO 2? 2NO + O 2 2NO 2 Example: Hw many grams f SO 3 are prduced when 55.3 grams f SO 2 reacts with an excess f O 2. Hmewrk: pg 311 #2-4 3
Sectin 9.3 Limiting Reactants and Percentage Yield Limiting reactant: the substance that determines the amunt f prduct that can be frmed by a reactin. Excess reactant: the substance that is nt cmpletely used up in a reactin. Keep in mind that the reactant that is present in the smaller amunt by mass r vlume is nt necessarily the limiting reactant. Determining Limiting Reactant What is the limiting reactant when 80.0g Cu reacts with 25.0g S? 2Cu(s) + S(s) Cu2S(s) Using the grams f reactant given, determine the mles f reactant. *Have Using the balanced chemical equatin, determine the mle rati. Multiply the mles f ne reactant by the mle rati t get the needed amunt f the ther reactant and cmpare t the amunt yu have. *When yu need mre than yu have, that is yur limiting reactant. *When yu have mre than yu need, that is yur excess reactant. Example: If 4.00 ml C 2 H 4 is reacted with 9.50 ml O 2, identify the limiting reactant and calculate the mles f excess reactant remaining. C 2 H 4 (g) + 3O 2 (g) 2CO 2 (g) + 2H 2 O(g) *Since we already have mles, we knw what we have, we just have t find what we need. Using a Limiting Reactant t Find Quantity f a Prduct What is the maximum number f grams f Cu 2 S that can be frmed when 45.0g Cu reacts with 24.0g S? 2Cu(s) + S(s) Cu 2 S(s) Find the limiting reactant and then use the mles f limiting reactant t calculate mles and then grams f Cu 2 S. Limiting Reactant is, s we use the given amunt f t find grams f Cu 2 S 4
Example Calculate the mles f I 2 prduced when 80.0g I 2 O 5 reacts with 28.0g CO. I 2 O 5 (g) + 5CO(g) 5CO 2 (g) + I 2 (g) Limiting Reactant is, s we use the given amunt f t find mles f I 2. We already have cnverted t mles in the first part, s start with mles. Hmewrk: page 321 #22-23 Percent Yield Theretical Yield: the maximum amunt f prduct that culd be frmed frm given amunts f reactants. Actual Yield: the amunt f prduct that actually frms when the reactin is carried ut in the labratry. Percent Yield: the rati f the actual yield t the theretical yield expressed as a percent. Percent Yield = actual yield x 100% units can be gram r ml, as theretical yield lng as they are the same. The percent yield is a measure f the f a reactin carried ut in the labratry. Factrs that can cause the actual yield t be less than the theretical yield: Calculating Theretical Yield What is the theretical yield in grams f CaO if 24.8g CaCO 3 is heated? CaCO 3 (s) CaO(s) + CO 2 (g) Calculate the theretical yield by cnverting the mass f the reactant t the mass f the prduct, using mlar mass and mle ratis. 5
Example What is the theretical yield f AlCl 3 when 3.00ml Al reacts with an excess f Cl 2? 2Al + 3Cl 2 2AlCl 3 Calculating Percent Yield What is the percent yield if 13.1g CaO is actually prduced when 24.8g CaCO 3 is heated? CaCO 3 (s) CaO(s) + CO 2 (g) Find the theretical yield (13.9g CaO) and plug bth the theretical and actual yields int the percent yield equatin. Example What is the percent yield if 2.85ml AlCl 3 is actually prduced when 3.00ml Al reacts with an excess f Cl 2? 2Al + 3Cl 2 2AlCl 3 Anther Way T Lk At It Think f percent yield as a grade n a test. The Theretical Yield is the number f pints pssible: 70 pts The Actual Yield is the number f pints yu earned: 64 pts Percent yield = 64/70 x 100% = 91% Hmewrk: page 318 #1 & 3 6