Answer to Question #263298 in Chemistry for nish

Question

Ethylene (C2H4 ), an important organic chemical, can be prepared by heating hexane (C6H14 ), at 800°C: C6H14 → C2H4 + other products If the yield of ethylene production is 42.5 %, what is the mass of hexane must be reacted to produce 481 g of ethylene

Answer 1

Let's assume that a complete and balanced equation for this reaction shows a 1:1 molar ratio of hexane to ethylene.

C₆H₁₄ → C₂H₄ + other products

First, we need to find the theoretical yield of ethylene.

Let "X" represent the theoretical yield of ethylene. Then 0.425 X = 481 grams of ethylene. Solving for "X", X = (481 g) / (0.425) , or X = 1,130 g rounded to three significant figures.

Next, we can use dimensional analysis to convert the theoretical yield of ethylene (1,120 g) to moles of ethylene, to moles of hexane, to grams of hexane. We will need the following equalities to make conversion factors:

1 mol C₂H₄ = 28.054 g C₂H₄

1 mol C₂H₄ = 1 mol C₆H₁₄

1 mol C₆H₁₄ = 86.178 g C₆H₁₄

[(1130 g C₂H₂)/1][(1 mol C₂H₄)/(28.054 g C₂H₄)][(1 mol C₆H₁₄)/(1 mol C₂H₄)][(86.178 g C₆H₁₄)/(1 mol C₆H₁₄)] = 3,470 g of C₆H₁₄ or 3.47 x 10³ g of C₆H₁₄ rounded to three significant figures.