C4H6(g)+2Cl(g)=C4H6Cl4(l) Calculate the mass of C4H6Cl4 that can be prepared from 35.0g of C4H6 and 105.0g of Cl2?

Question

Not sure where to start. Step by step please

Answer 1

Always start with a correctly balanced equation for the reaction taking place.

C₄H₆(g) + 2Cl₂(g) ==> C₄H₆Cl₄(l) ... balanced equation (note the equation given in the problem is incorrect)

Next, find the limiting reactant. Any time you are given mass or moles of BOTH reactants, you must find the limiting reactant before proceeding.

Limiting reactant:

For C4H6:

35.0 g C₄H₆ x 1 mol C₄H₆/54.09 g x 1 mol C₄H₆Cl₄/1 mol C₄H₆ x 125 g/mol C₄H₆Cl₄ = 80.88 g C₄H₆Cl₄

For Cl₂:

105.0 g Cl₂ x 1 mol Cl₂/70.91 g x 1 mol C₄H₆Cl₂/2 mol Cl₂ x 125 g/mol C₄H₆Cl₂ = 92.55 g C₄H₆Cl₄

LIMITING reactant = C₄H₆

So, the mass of C₄H₆Cl₂ that can be prepared is limited by the mass (moles) of C₄H₆ and the answer would be 80.8 g C₄H₆ (to 3 significant figures).