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.
Answer:
Number of protons = 52, Number of electrons = 52, Number of neutrons = 76
Explanation:
The balanced chemical reaction is:
Zn + 2AgNO3 = Zn(NO3)2 + 2Ag
To determine the amount of the reactant left, we have to determine which is the limiting and the excess reactant. We do as follows:
5.65 g Zn ( 1 mol / 65.38 g) = 0.09 mol Zn
21.6 g AgNO3 (1 mol / 169.87 g) = 0.13 mol AgNO3
The limiting reactant would be silver nitrate since it is consumed completely in the reaction. The excess reactant would be zinc.
Excess Zinc = 0.09 mol Zn - (0.13 / 2) mol Zn = 0.025 mol Zn left
The chemical reaction would be written as follows:
2Al + 3Cl2 = 2AlCl3
We are given the amount of aluminum to be used in the reaction. This will be the starting point of the calculations. We do as follows:
19.0 g Al ( 1 mol / 29.98 g ) ( 2 mol AlCl3 / 2 mol Al ) = 0.63 mol AlCl3
The formula of Density is known as:
Density = mass/volume
Units: g/cm^3
Hope this helps!
