Question

Limiting Reactant, Theoretical Yield and Percent Yield 2. Pentane combusts with en to form carbon dioxide and water by the following reaction: CsHız(1) + 8 O2(g) → 5 CO2(g) + 6H2O(g) a. If 8.00 g of pentane is mixed with 10.0 g of oxy compare it to either of the products). he is mixed with 10.0 g of oxygen, which is the limiting reactant? (Hint: b. What is the theoretical yield (in grams) of carbon dioxide and water for this reaction

Answer 1

2.a) CG H 1 2 (d) + 8 O₂(g) - 5 CO₂(g) + 6 H₂O (9) I mol of Cf H 2 neacts with 8 mol of one Molaro mass of CG H2 = 72 g/mol Molaro mass of O2 = 32 g/mol 72 g of CG H 12 reacts with (8x32) g = 256 g of Oz 8 g of His Beacts with 256% 8 a ? 72 g = 28.44 g of . s limited by Limiting neagent is the substance which is totally consumed when the chemical reaction is complete. .. oxygen) is the limiting reactant, 6 Amount of product formed is limited to limiting rocagent. Molaro mass of CO2 = 44 g/mol Molaro mass of H₂O = 18 g/mol 8 mod of or produces 5 mol of co2 (8+82) g = 256 g of O2 produces (5 x 44) g = 220 g of CO2 and log of Oz produces 220x10 g = 8.59 g of con Again, 256 g of O2 produces (6x 18 ) g = 108 g of H₂o - log of O2 produces 108 x 10 g = 4.22 g of Ho" ... Theoretical yield of co2 = 18.59 g Theoretical yield of H₂O = (4.22 g M