Answer:
There is a mass of 154 Grams of Carbon Dioxide.
Explanation:
One mole is equal to 6.02 × 10^23 particles.
This means we have 1.05 X 10^24 total particles of Ethane.
Each ethane particle contains 2 carbon atoms.
If every particle of ethane is burned, we will end up with 2.10 x 10^24 molecules of Carbon Dioxide (Particles of Methane x 2, since each Methane particle contains 2 carbon atoms)
Carbon Dioxide has a molar mass of 44.01 g/mol
So if we take our amount of Carbon Dioxide molecules and divide it by 1 mole, ((2.10 x 10^24)/(6.02 x 10^23) = 3.49) we find that we have 3.49 moles of Carbon Dioxide.
Now all we need to do is multiply our moles of carbon dioxide(3.49) by it's molar mass(44.01) while accounting for significant digits.
What you should end up with is 154 Grams of Carbon Dioxide.
Hope this helps (And more importantly I hope I didn't make any errors in my math lol)
As a side note this is all assuming that this takes place at STP conditions.
2. <span>High pressure and low temperature
Hope this helps </span>
Answer:
588.2 mL
Explanation:
- FeSO₄(aq) + 2KOH(aq) → Fe(OH)₂(s) + K₂SO₄(aq)
First we <u>calculate how many Fe⁺² moles reacted</u>, using the given <em>concentration and volume of FeSO₄ solution</em> (the number of FeSO₄ moles is equal to the number of Fe⁺² moles):
- moles = molarity * volume
- 187 mL * 0.692 M = 129.404 mmol Fe⁺²
Then we convert Fe⁺² moles to KOH moles, using the stoichiometric ratios:
- 129.404 mmol Fe⁺² *
= 258.808 mmol KOH
Finally we<u> calculate the required volume of KOH solution</u>, using <em>the given concentration and the calculated moles</em>:
- volume = moles / molarity
- 258.808 mmol KOH / 0.440 M = 588.2 mL
