The reaction scheme goes as follows: H2PtCl6 CH3--(CH2)5--CH==CH2 <-----------------> CH3--(CH2)2--CH2--CH3 1-Octene NaBH4 Octane Ethanol 6M HCl (diluted) Givens: 120µL = 0.12mL of 1-Octene 50µL = 0.050mL of H2PtCl6 1mL of Ethanol 125µL = 0.125mL of NaBH4 100µL = 0.100mL of diluted 6M HCl -------------------- My attempt: I know that we have to convert what we have to moles (using mmols) 0.12mL of 1-Octene (using the density=0.72g/mL) = 0.76 mmol 50µL H2PtCl6 = 0.234 mmol 1mL Ethanol = 17.1 mmol 125µL NaBH4 = 3.54 mmol we can ignore the HCl because it's diluted, right? (also since we don't know the % of actual HCl dilution, so we can't use the molarity to get moles) Since we are going from 1-Octene to Octane, we have gained an extra H+ and lost a double bond. So essentially, the H2PtCl6 catalyst must have given off a H+ (since Pt metal targets alkenes). My dilemma is that I'm unsure on how to find the actual limiting reactant, and what information I should use. There's too much given; but from my introspection, we only need 1-Octene and the Pt catalyst,