Chemical reactios ad equatios Example: Coservatio of mass Cosider the followig reactio for the electrolysis of water: electricity 2 H 2 2 H 2 + 2 If 22.4 g of hydroge ad 177.6 g of oxyge are formed, how may grams of water reacted? Coservatio of mass mass of reactats = mass of products mass of H 2 = mass of H 2 + mass of 2 mass of H 2 = 22.4 g + 177.6 g mass of H 2 = 200.0 g I a balaced chemical equatio, the total umber of atoms of each elemet must be the same o both sides of the equatio -- you ca thik of this as applyig accoutig priciples to chemistry Equatio: Review: Balaced chemical equatios C 3 H 8 (g) + 5 2 (g) 3 C 2 (g) + 4 H 2 (g) propae oxyge carbo water dioxide Calculatios from chemical equatios What iformatio about a chemical reactio ca you get from a balaced chemical equatio? C 3 H 8 (g) + 5 2 (g) 3 C 2 (g) + 4 H 2 (g) If you are give the amout of ay reactat or product ivolved i the reactio: 3 carbo atoms, 8 hydroge atoms, 10 oxyge atoms 3 carbo atoms, 8 hydroge atoms, 10 oxyge atoms you ca calculate the amouts of all the other reactats ad products that are cosumed or produced i the reactio A balaced chemical equatio is a expressio of the Law of Coservatio of Mass Matter ca ot be created or destroyed -- it ca oly shift from oe form to aother I a chemical reactio, o atoms are created or destroyed -- they are just recombied to form ew substaces
Bakig recipe aalogy Bakig recipe aalogy 1 bag flour + 1 carto milk + 6 eggs 24 pacakes 1 bag flour + 1 carto milk + 6 eggs 24 pacakes If you use 2 bags of flour, how may pacakes will you get? If you wat to make 36 pacakes, how may eggs will you eed to use? (2 bags flour) 24 pacakes 1 bag flour X pacakes = (2 bags flour) 2 bags flour (36 pacakes) 6 eggs 24 pacakes X eggs = (36 pacakes) 36 pacakes 48 pacakes = X 9 eggs = X Back to chemical reactios Example: Copper ad oxyge, whe heated, combie to form copper (II) oxide + 2 + (1) 2 2 If you react 6 atoms of, how may molecules of will you get? 2 molecules X molecules (6 atoms ) = (6 atoms ) 2 atoms 6 atoms Remember: I a balaced chemical equatio, coefficiets idicate the relative umbers of formula uits ad relative umbers of moles of each reactat / product + 2 + (1) 2 2 2 atoms 1 2 molecule 2 molecules 6 molecules = X
Remember: I a balaced chemical equatio, coefficiets idicate the relative umbers of formula uits ad relative umbers of moles of each reactat / product + 2 + (1) 2 2 2 moles 1 mole 2 2 moles Calculatios from chemical equatios If you kow the amout of ay reactat or product ivolved i the reactio: you ca calculate the amouts of all the other reactats ad products that are cosumed or produced i the reactio C 3 H 8 (g) + 5 2 (g) BUT REMEMBER! 3 C 2 (g) + 4 H 2 (g) The coefficiets i a chemical equatio provide iformatio NLY about the proportios of MLES of reactats ad products give the umber of moles of a reactat/product ivolved i a reactio, you CAN directly calculate the umber of moles of other reactats ad products cosumed or produced i the reactio 2*(6.022 x 10 23 atoms/mol) = 1.204 x 10 24 atoms 1*(6.022 x 10 23 molecules/mol) = 6.022 x 10 23 molecules 2 2*(6.022 x 10 23 molecules/mol) = 1.204 x 10 24 molecules give the mass of a reactat/product ivolved i a reactio, you ca NT directly calculate the mass of other reactats ad products cosumed or produced i the reactio Mole - mole calculatios Give: A balaced chemical equatio A kow quatity of oe of the reactats/product (i moles) Calculate: The quatity of oe of the other reactats/products (i moles) Example: Mole - mole calculatios How may moles of ammoia are produced from 8.00 mol of hydroge reactig with itroge? Equatio: 3 H 2 + N 2 2 NH 3 Use ratio betwee coefficiets of substaces A ad B from balaced equatio Mole ratio betwee ukow substace (ammoia) ad kow substace (hydroge): 2 moles NH 3 3 moles H 2 substace A substace B 2 moles NH 3 ( 8.00 moles H 2 ) = ( 8.00 moles H 2 ) 8.00 moles H 2 3 moles H 2 = 5.33 moles NH 3
Mole - mole calculatios Mole - mole calculatios Give the balaced equatio: Give the balaced equatio: K 2 Cr 2 7 + 6 KI + 7 H 2 S 4 Cr 2 (S 4 ) 3 + 4 K 2 S 4 + 3 I 2 + 7 H 2 K 2 Cr 2 7 + 6 KI + 7 H 2 S 4 Cr 2 (S 4 ) 3 + 4 K 2 S 4 + 3 I 2 + 7 H 2 Calculate: a) The umber of moles of potassium dichromate (K 2 Cr 2 7 ) required to react with 2.0 mol of potassium iodide (KI) Calculate: b) The umber of moles of sulfuric acid (H 2 S 4 ) required to produce 2.0 moles of iodie (I 2 ) Mole ratio betwee the ukow substace (potassium dichromate) ad the kow substace (potassium iodide): 1 mol K 2 Cr 2 7 6 mol Kl Mole ratio betwee the ukow substace (sulfuric acid) ad the kow substace (iodie): 7 mol H 2 S 4 3 mol l 2 1 mol K 2 Cr 2 7 ( 2.0 mol KI ) = ( 2.0 mol KI ) 2.0 mol Kl 6 mol Kl 7 mol H 2 S 4 ( 2.0 mol l 2 ) = ( 2.0 mol l 2 ) 2.0 mol l 2 3 mol l 2 = 0.33 mol K 2 Cr 2 7 = 4.7 mol H 2 S 4 Give: Mass - mass calculatios A balaced chemical equatio A kow mass of oe of the reactats/product (i grams) Mass - mass calculatios How may grams of itric acid are required to produce 8.75 g of diitroge mooxide (N 2 )? Calculate: The mass of oe of the other reactats/products (i grams) 4 Z (s) + 10 HN 3 (aq) 4 Z(N 3 ) 2 (aq) + N 2 (g) + 5 H 2 (l) Grams of substace A Grams of substace B Step 1: Covert the amout of kow substace (N 2 ) from grams to moles Use molar mass of substace A substace A Use ratio betwee coefficiets of substaces A ad B from balaced equatio substace B Use molar mass of substace B Molar mass N 2 : ( 2 x 14.01 g/mol ) + 16.00 g/mol = 44.02 g/mol 8.75 g N 2 ( 1 mol N 2 / 44.02 g N 2 ) = 0.199 mol N 2
Mass - mass calculatios How may grams of itric acid are required to produce 8.75 g of diitroge mooxide (N 2 )? 4 Z (s) + 10 HN 3 (aq) 4 Z(N 3 ) 2 (aq) + N 2 (g) + 5 H 2 (l) Step 2: Determie the umber of moles of the ukow substace (HN 3 ) required to produce the umber of moles of the kow substace (0.199 mol N 2 ) Mole ratio betwee the ukow substace (itric acid) ad the kow substace (diitroge mooxide): 10 mol HN 3 1 mol N 2 10 mol HN 3 ( 0.199 mol N 2 ) 0.199 mol N 2 = ( 0.199 mol N 2 ) 1 mol N 2 Mass - mass calculatios How may grams of itric acid are required to produce 8.75 g of diitroge mooxide (N 2 )? 4 Z (s) + 10 HN 3 (aq) 4 Z(N 3 ) 2 (aq) + N 2 (g) + 5 H 2 (l) Step 3: Covert the amout of ukow substace (1.99 moles HN 3 ) from moles to grams Molar mass HN 3 : 1.008 g/mol + 14.01 g/mol + ( 3 x 16.00 g/mol ) = 63.02 g/mol 1.99 mol HN 3 ( 63.02 g HN 3 / 1 mol HN 3 ) = 125 g HN 3 = 1.99 mol HN 3 Mass - mass calculatio: Aother example How may grams of carbo dioxide are produced by the complete combustio of 100. g of petae (C 5 H 12 )? C 5 H 12 (g) + 8 2 (g) 5 C 2 (g) + 6 H 2 (g) Step 1: Covert the amout of kow substace (C 5 H 12 ) from grams to moles Molar mass C 5 H 12 : ( 5 x 12.01 g/mol ) + ( 12 x 1.008 g/mol ) = 72.15 g/mol 100. g C 5 H 12 ( 1 mol C 5 H 12 / 72.15 g C 5 H 12 ) = 1.39 mol C 5 H 12 Mass - mass calculatio: Aother example How may grams of carbo dioxide are produced by the complete combustio of 100. g of petae (C 5 H 12 )? C 5 H 12 (g) + 8 2 (g) 5 C 2 (g) + 6 H 2 (g) Step 2: Determie the umber of moles of the ukow substace (C 2 ) required to produce the umber of moles of the kow substace (1.39 mol C 5 H 12 ) Mole ratio betwee the ukow substace (carbo dioxide) ad the kow substace (petae): 5 mol C 2 1 mol C 5 H 12 5 mol C 2 ( 1.39 mol C 5 H 12 ) = ( 1.39 mol C 5 H 12 ) 1.39 mol C 5 H 12 1 mol C 5 H 12 = 6.95 mol C 2
Mass - mass calculatio: Aother example How may grams of carbo dioxide are produced by the complete combustio of 100. g of petae (C 5 H 12 )? C 5 H 12 (g) + 8 2 (g) 5 C 2 (g) + 6 H 2 (g) Chapter 6 Problems: Homework assigmet 6.63, 6.64, 6.66, 6.67, 6.68 6.69 Step 3: Covert the amout of ukow substace ( 6.95 moles C 2 ) from moles to grams Molar mass C 2 : 12.01 g/mol + ( 2 x 16.00 g/mol ) = 44.01 g/mol = 6.95 mol C 2 ( 44.01 g C 2 / 1 mol C 2 ) = 306 g C 2