Answer:
336.6 grams of CO₂ and 183.6 grams of H₂O are formed from 2.55 moles of propane.
Explanation:
In this case, the balanced reaction is:
C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reactant and product participate in the reaction:
- C₃H₈: 1 mole
- O₂: 5 moles
- CO₂: 3 moles
- H₂O: 4 moles
Being the molar mass of each compound:
- C₃H₈: 44 g/mole
- O₂: 16 g/mole
- CO₂: 44 g/mole
- H₂O: 18 g/mole
Then, by stoichiometry, the following quantities of mass participate in the reaction:
- C₃H₈: 1 mole* 44 g/mole= 44 grams
- O₂: 5 moles* 16 g/mole= 80 grams
- CO₂: 3 moles* 44 g/mole= 132 grams
- H₂O: 4 moles* 18 g/mole= 72 grams
So you can apply the following rules of three:
- If by stoichiometry 1 mole of C₃H₈ forms 132 grams of CO₂, 2.55 moles of C₃H₈ how much mass of CO₂ will it form?

mass of CO₂= 336.6 grams
- If by stoichiometry 1 mole of C₃H₈ forms 72 grams of H₂O, 2.55 moles of C₃H₈ how much mass of H₂O will it form?

mass of H₂O= 183.6 grams
<u><em>336.6 grams of CO₂ and 183.6 grams of H₂O are formed from 2.55 moles of propane.</em></u>
Answer:
The answer to your question is 1.11 M
Explanation:
Data
volume 1 = 287 ml
concentration 1 = 1.6 M
volume 2= 412 ml
concentration 2 = ?
Formula
Volume 1 x concentration 1 = Volume 2 x concentration 2
Solve for concentration 2
concentration 2 = (volume 1 x concentration 1) / volume 2
Substitution
concentration 2 = (287 x 1.6) / 412
Simplification
concentration 2 = 459.2 / 412
Result
concentration 2 = 1.11 M
Answer:
The part of the experiment that is set to compare data
Explanation:
The controlled variable is what stays constant throughout an experiment. You use the controlled variable to compare the new data to see what happened during whatever reaction you could say.
Answer:
D:Nuclear fusion plays an important role in making elements that are heavier than helium
Explanation:
just took the test