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Stoich!

Stoich!. Finishing S’Mores lab and movin’ on to the good stuff!. 1 Gg ‘atom’ = 0.357 grams 1 Cc ‘atom’ = 0.510 grams 1 Mm ‘atom’ = 0.616 grams. You have 64 ‘atoms’ of Gg….how many S’mores ( Gg 2 Cc 3 Mm)?. 2 Gg + 3 Cc + Mm  Gg 2 Cc 3 Mm. 64 Gg x 1 S’more = 32 S’mores. 2 Gg.

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Stoich!

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  1. Stoich! Finishing S’Mores lab and movin’ on to the good stuff!

  2. 1 Gg ‘atom’ = 0.357 grams 1 Cc ‘atom’ = 0.510 grams 1 Mm ‘atom’ = 0.616 grams You have 64 ‘atoms’ of Gg….how many S’mores ( Gg2Cc3Mm)? 2 Gg + 3 Cc + Mm  Gg2Cc3Mm 64 Gg x 1 S’more = 32 S’mores 2 Gg

  3. Question 2 1 Gg ‘atom’ = 0.357 grams 1 Cc ‘atom’ = 0.510 grams 1 Mm ‘atom’ = 0.616 grams 2 Gg + 3 Cc + Mm  Gg2Cc3Mm You have 54.97 g Gg…what mass of S’mores could you make? a) Mass of Gg = 0.357 g 4 sig figs 154 ‘atoms’ Gg b) How many atoms Gg? Total mass = 54.97 g Gg = 153.978 ‘atoms’ Gg Mass 1 Gg 0.357 g per Gg 3 sig. figs c) 154 Gg x 1 S’more = 77.0 S’mores 2 Gg d) Total mass of S’mores? 1st find mass of one S’more: 2(.357g) + 3(.510g) + .616g = 2.86 g (is the weight of 1 S’more) 220. g 77.0 S’mores x 2.86 g per S’more = 220.22g

  4. 2 Gg + 3 Cc + Mm  Gg2Cc3Mm 1) How many moles of Mm would be required to react completely with 4.5 mole Cc? 4.5 mole Cc x 1 mole Mm = 1.5 mol Mm 3 mol Cc 2) Grams of Cc ? Start with 178.5 g Gg. 178.5 g Gg x 1 Gg x 3 mol Cc x .510 g Cc = 382.5 g Cc .357 g Gg 2 mol Gg 1 mol Cc 3) 2(.357g) + 3(.510g Cc) + .616g Mm = 2.86 g S’mores 4) % Gg = 2(.357g) x 100% = 25.0% Gg 2.86 g % Cc = 3(.510g) x 100% = 53.4% Cc 2.86 g % Mm = .616 g x 100% = 21.5 % Mm 2.86 g

  5. S’Mores Lab…then No (s)’ more!  2 Gg + 3 Cc + Mm  Gg2Cc3Mm 5) 193 Cc X1 Cc X1 S’More X 2.86 g S’Mores = 361 g S’Mores .510 g Cc 3 mol Cc 1 S’More 6) 6 mol Al X1 mol Al2S3 = 3 mol Al2S3 2 mol Al 96.18 g S X 1 mole SX1 mole Al2S3X (2(26.982) + 3(32.065)) g Al2S3 32.065 g S 3 mole S 1 mole Al2S3 =150.1 g Al2S3

  6. Stoichiometry (Stoyk-e-ometry) Definition: Using an amount of one substance to predict how much of a reactant will be used or a product will be produced. Reactants  Products Steps: • 0. Balance the equation if it isn’t already balanced. • Convert mass  moles • Convert moles of ‘known’  moles of ‘unknown’ (whatever you’re trying to find). Use the mole ratios (coefficients) from the balanced equation. • Convert moles of ‘unknown’  mass ‘unknown’ How many grams of H2O will react with 73.4 grams of CO2? Example: 3CO2 + 4H2O  C3H8 + 5O2 73.4g CO2x1 mol CO2 x4 mol H2O X18.0148 g H2O = 40.1 g H2O 44.009 g CO2 3 mol CO2 1 mol H2O Final Answer!

  7. Problem: How many grams of Cl2 will react with 100. grams of Fe? Simply Stoich-ED! 1) 2 Fe + 3Cl2 2FeCl3 2) 100. g Fe X1 mole Fe = 1.790… mol Fe 55.845 g Fe 3) 1.790.. mol Fe X3 mol Cl2= 2.686… mol Cl2 2 mol Fe 4) 2.686… mol Cl2X2(35.453) g Cl2 = 190. g Cl2 1 mol Cl2 Another way to write this: 100. g Fe X 1 mole FeX3 mol Cl2 X2(35.453)g Cl2= 190. g Cl2 55.845 g 2 mol Fe 1 mol Cl2

  8. Got help? Try this: Helpful Links: http://science.widener.edu/svb/tutorial/rxnsgramstogramscsn7.html

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