is ethylene was mixed with oxygen in a combustion reaction, how much carbon dioxide would be produced from 20g of ethylene?

Answer 1

Let's look at the equation for the reaction taking place:

ethylene = C₂H₄

C₂H₄ + 3O₂ ===> 2CO₂ + 2H₂O ... balanced equation for combustion of ethylene

assuming O2 is present in excess,...

20 g C₂H₄ x 1 mol C₂H₄ / 28 g x 2 mol CO₂ / mol C₂H₄ x 44 g CO₂ / mol CO₂ = 63 g CO₂ produced (2 s.f.)

Answer 2

Step 1: Form a chemical equation

In this problem, ethylene (C₂H₄) reacts with oxygen (O_2). As a rule of thumb, whenever you have a hydrocarbon reacting with oxygen in a combustion reaction, the products of the reaction will always be water (H₂O) and carbon dioxide (CO₂). Therefore we get the following reaction equation...

C₂H₄ + O₂ --> H₂O + CO₂

Step 2: Balance the chemical equation

Whenever balancing chemical equations, you ultimate goal is to get the same number of atoms of each element on both sides of the reaction arrow. We are able to balance chemical reactions by adding "coefficients". Starting out, on the left side of the arrow, we have 2 C, 4 H, and 2 O. On the right we have 1 C, 2 H, and 3 O. It is easiest to save what I like to refer to as "the most complex element" last. What I mean by "most complex" is the element that show up the most in the chemical equation. For this equation, the most complex element is oxygen (since it shows up in 3 places in the equation). So, balance the oxygen atoms last. Below is the balanced equation

1 C₂H₄ + 3 O₂ --> 2 H₂O + 2 CO₂

Step 3: Unit Conversions (i.e. stoichiometry)

Math in chemistry is usually starting with some number with a specific unit and converting it into something else using a series of conversion factors. "Conversion factor" is a simply a fancy term for a "fraction". The conversion factors most commonly used are molar mass, Avogadro’s number, prefixes such as kilo- or milli-, and mole ratios from balanced chemical reaction equations. 

Begin by writing down what you are given, 20 grams ethylene, and put it over 1 to form the first fraction. From there, the first conversion factor is the molar mass of ethylene (found using the periodic table). The second conversion factor is from the coefficients of the balanced chemical equation. The third conversion factor is the molar mass of CO₂ (found using the periodic table). The final step is to multiply these fractions by one another. In my tutoring sessions, I teach student a simple technique that makes this process very easy.

(20 g ethylene/1) * (1 mol ethylene/28.06 g ethylene) * (2 mol CO₂/1 mol ethylene) * (44.01 g CO₂/ 1 mol CO₂) = 62.74 g CO₂

Answer: 62.74 g CO₂