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.
Find the mass of C in the 2.657 g CO2:
(2.657 g CO2) / (44.01 g/mol) = 0.06037 mol CO2
Since each mole of CO2 also has 1 mole of C, this is equivalent to 0.06037 mol C.
Find the mass of H in the 1.089 g H2O:
(1.089 g H2O) / (18.02 g/mol) = 0.06043 mol H2O
Since 1 mol H2O has 2 mol H, this is equivalent to (0.06043)*2 = 0.1209 mol H.
Taking the ratio of H to C: 0.1209 / 0.06037 = 2.002 ~ 2
Therefore, the empirical formula of isobutylene is CH2.
1 to 1. Most small atoms have the same number of protons and neutrons
2H2 (g) + O2 (g) -->2H2 O(g)
mole ratio of H2:O2=2:1
7.25/2=3.625
An extinction event is a widespread and rapid decrease in the biodiversity on Earth. Such an event is identified by a sharp change in the diversity and abundance of multicellular organisms. It occurs when the rate of extinction increases with respect to the rate of speciation. -Wikipedia
