Question

For the reaction of ethene, C2H4(g), with oxygen(g) to form carbon dioxide(g) and water(g), how many grams of carbon dioxide could be produced from 2.0 g of ethene and 2.9 g of oxygen?

Answer 1

well I guess since no one else wants to (or can't) answer it, I'll take a stab (:

C2H4 (g) + O2 (g) --> CO2 (g) + H2O (g)

The first step we need to do is balance the reaction. Can you do that?

Answer 2

C2H4 + 3O2 --> 2CO2 + 2H2O

Answer 3

Great! Now we need to convert the grams to moles.

Do you know how to find the molar mass of ethene and the molar mass of oxygen gas (O2)?

Answer 4

28.05 for ethene and for oxygen its about 32

Answer 5

You're doing wonderful!

To convert from grams to moles, we just divide by the molar mass

2.0 g Ethene * 1 mol/28.05 g =

2.9 g O2 * 1 mol/32 g =

Answer 6

1: 0.071301248
2: 0.090625

Answer 7

So we have 0.0713 moles of ethene and 0.0906 moles of oxygen

We have to go back to our chemical reaction. Remember, everything is molar ratio.

\(\text{C}_2\text{H}_4 \text{(g)} + 3 \text{O}_2 \text{(g)} \to 2 \text{CO}_2 \text{(g)} + 2 \text{H}_2\text{O (g)}\)

1 mole of C2H4 will react 3 mole of O2 to produce 2 moles of CO2 and 2 moles of H2O

The mole to mole ratio is
1:3:2:2

So if we have 0.0713 moles of ethene, how many moles of O2 would react with it?

Answer 8

remember, it's a 1:3 ratio

If we have 1 mole of ethene, we need 3 moles of oxygen

so if we have 0.0713 moles of ethene, how many moles of oxygen would we need?

Answer 9

oh um wow, i uh idk

Answer 10

To make this smoothie for every 1 apple, I need 3 bananas

Similarly,
I have 0.0713 ethene, so how much oxygen do I need?

(just multiply by 3 because the ratio is 1 to 3)

Answer 11

oh ok

Answer 12

so 0.2139?

Answer 13

yes, now remember

\(\color{#0cbb34}{\text{Originally Posted by}}\) @person321
1: 0.071301248
2: 0.090625
\(\color{#0cbb34}{\text{End of Quote}}\)

we said there was 0.0906 moles of oxygen

if we want to use up all 0.0713 moles of ethene, we need 0.2139 moles of oxygen

do we have enough oxygen to use up all 0.0713 moles of ethene?

Answer 14

we only have 0.0906 moles of oxygen

we need 0.2139 moles of oxygen

do we have enough?

Answer 15

no we dont

Answer 16

This makes oxygen our limiting reagent. This is the reactant that we will use all up.

So now we have to find out how much ethane will get used if we use up all the oxygen

Following along still?

Answer 17

yeah

Answer 18

Good

so remember how it's all in a ratio

1 to 3 to 2 to 2

last time we were going from ethene to oxygen which is 1:3
but this time, we're going the other way around
so instead of multiplying by 3, we're dividing by 3

we have 0.0906 moles of O2 and we're going to use it all up

how many moles of ethene will react with 0.0906 moles of O2?

(divide by 3)

Answer 19

0.0302

Answer 20

and we had calculated initially that we have 0.0713 moles of ethene

and so we're only using up 0.0302 moles which means it all works!
(the excess can't do anything because there's no more oxygen for it to react with)

Answer 21

now we need to calculate how much of the product we are forming

Answer 22

ok

Answer 23

we just have to figure out how much product we're making

Answer 24

remember how I said everything is in moles to moles ratio

so we had

1 : 3 : 2 : 2

so

if we're using up 0.0302 ethene with 0.0906 moles of O2
how much of CO2 and H2O are we going to form?

The ratio is 1 mole of ethene will form 2 moles of CO2 and H2O
so if we use 0.032 moles of ethene, how many moles of CO2 and H2O are we forming?

(multiply by 2)

Answer 25

Well, you went offline and wow I've been on this post for an hour now.

We're almost done. Here are the last steps:

Once you multiply 0.032 by 2, you'll get the number of moles for both CO2 and H2O

Next, you have to find the molar mass of CO2 and the molar mass of H2O

Multiply the number of moles with their respective molar mass, and you'll get how much product is formed in grams

Answer 26

I'll still be around if you want me to check your work :)