limiting reactant: excess reactant:

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Presentation transcript:

limiting reactant: excess reactant: How many complete cars can be made from these available parts? 1 Complete Race Car only 2 Engines available: In chemistry, the amount of product that can be made is limited by the amount of one reactant. Image from POGIL (Flinn HSPI) Limiting Reactants limiting reactant: (completely consumed) determines the amount of product formed excess reactant: (left over)

? N2(g) + 3 H2(g) 2 NH3(g) excess limiting “left over” “runs out” 0 H2 molecules “left over” “runs out” 3 mol N2 is enough to make 6 mol NH3 , but… H2 runs out after making only 2 mol NH3 .

Mg + 2 HCl → MgCl2 + H2 Determining Limiting Reactant (with Math not Pics) Determining Limiting Reactant convert reactant A to reactant B to compare If available < needed (limiting) If available > needed (excess) Which is limiting? Mg + 2 HCl → MgCl2 + H2 48.6 g Mg 110. g HCl 1 mol Mg 24.31 g Mg 2 mol HCl 1 mol Mg 36.46 g HCl 1 mol HCl 48.6 g Mg x 146 g HCl x x = (110 g HCl) available < needed (146 g HCl) limiting HCl is limiting 1 mol HCl 36.46 g HCl 1 mol Mg 2 mol HCl 24.31 g Mg 1 mol Mg 110. g HCl x x x = 36.7 g Mg consumed 3

Mg + 2 HCl → MgCl2 + H2 Determining Limiting Reactant: (with Math not Pics) Determining Limiting Reactant: convert reactant A to reactant B to compare If available < needed (limiting) If available > needed (excess) Which is limiting? Mg + 2 HCl → MgCl2 + H2 48.6 g Mg 110 g HCl initial –36.7 g –110 g consumed (reacted) 11.9 g Mg 0 g HCl left over excess limiting limiting 1 mol HCl 36.46 g HCl 1 mol Mg 2 mol HCl 24.31 g Mg 1 mol Mg 110 g HCl x x x = 36.7 g Mg consumed 4

Mg + 2 HCl → MgCl2 + H2 2nd Way (kind of) 48.6 g Mg 110. g HCl 1 mol MgCl2 2 mol HCl 143 g MgCl2 produced 1 mol HCl 36.46 g HCl 95.21 g MgCl2 1 mol MgCl2 110. g HCl x x x = 1 mol MgCl2 1 mol Mg 190. g MgCl2 produced 48.6 g Mg x 1 mol Mg 24.31 g Mg 95.21 g MgCl2 1 mol MgCl2 x x = HCl is limiting 5

C2H4 + 3 O2 → 2 CO2 + 2 H2O Determining Product with LR: How many grams CO2 can be formed from 2.70 mol C2H4 and 9.30 mol O2? C2H4 + 3 O2 → 2 CO2 + 2 H2O 3 mol O2 1 mol C2H4 2.70 mol C2H4 x = 8.10 mol O2 (9.30 mol O2) available > needed (8.10 mol O2) O2 is excess C2H4 is limiting limiting 2 mol CO2 1 mol C2H4 44.01 g CO2 1 mol C2H4 238 g CO2 produced 2.70 mol C2H4 x x = 6

Quick Quiz! 1. How many grams of H2O can be formed from 64 g O2 and 3 mol H2? 2 H2 + O2  2 H2O 90.1 g H2O 72.1 g H2O 54.1 g H2O 36.0 g H2O 2 mol H2O 2 mol H2 18.02 g H2O 1 mol H2O 3 mol H2 x x = 1 mol O2 32 g O2 2 mol H2 1 mol O2 64 g O2 x x = 4 mol H2 (3 mol H2) available < needed (4 mol H2) H2 is limiting