1. Given: mass = 0.334g of C2H6, CAtomicWeight = 12.01, HAtomicWeight = 1.008
Required:
a. Molar Mass of Ethane
b. Number of molecules of C2H6 in 0.334g of C2H6
Formula:
a. (n = (2 x CAtomicWeight) + (6 x HAtomicWeight)
b. mass x (1 mole / a ) x (6.022x1023 molecules / 1 mole)
Solution:
a. Molar Mass of Ethane
n = (2 x 12.01) + (6 x 1.008)
n = 24.02 + 6.0498
n = 30.0698
b. Number of molecules
n = 0.334 x (1 mole / 30.0698 g) x (6.022x1023 molecules / 1 mole)
n = 6.689x1021 molecules
FINAL Answer: The are 6.689x1021 molecules in 0.334g of Ethane
How many molecules are present in 42.0g of Cl2
12 molecules of carbon 24 of hydrogen 12 of oxygen
2.01 mol
0.175 X Avogadro's Number = about 1.05 X 1023.
? molc' MgBr2 = (13.77lbs)(453.59g/1lbs)(1 mol/184.11g)(6.02e23 molc'/1 mol) 13.77 lbs =2.04e25 molecules
mol. mass of ethane = 2x12.01+ 6x1.008= 30.01g/mol # of mole in 0.334 of ethane= 0.334/30.01= 0.01mol
Ethane does not have any molecule of carbon dioxide. However when ethane undergoes combustion then two molecules of carbon dioxide are formed (as ethane contains two carbon atoms).
The number of molecules in 15 g ethane is approx. 3,011.10e23.
Ethane is C2H6, with a molar mass of 30g/mol. 60g of ethane is 2mol. Avogadro's number (# of particles in a mole) = 6.02x1023. Avogadro x 2mol = 1.2x1024 ethane molecules.
at STP 1 mole occupies 22.4 litres. 64.28 / 22.4 is 2.8696428 moles. Multiply this by avagadro's constant (6.022*10^23) gives 1.7281x10^24 molecules
30 g of ethane will have 6.023 x 1023 molecules of ethane So, 5.5 g will have 1.104 x 1023 molecules of ethane Since there are two carbon atoms, in one molecule of ethane, 5.5 g of ethane will have 2.208 x 1023 atoms of carbon.
I am presuming that you are asking how many carbon dioxide (CO2) molecules will be formed when two ethane (C2H6) molecules burn in a plentiful supply of oxygen. The following balanced equation is that of the burning of ethane in a plentiful supply of oxygen: C2H6 + 3.5O2 -----> 2CO2 + 3H2O. The number before each of the molecules in the equation tells us how many there are in this reaction. This means that for every one ethane molecule that is burnt, two carbon dioxide molecules are produced.
You think probable to polymerization.
Only sigma bonds are present in ethane. There is one carbon-carbon sigma bond and six carbon-hydrogen sigma bonds in C2H6.
The reaction is :- 2C2H6 + 7O2 ----------> 4CO2 + 6H2O When one mole ethane is combusted 7/2 moles of oxygen are used. When 3 moles of ethane are combusted 3 x 7/2 moles of oxygen used. No. of oxygen molecules consumed =6.022 x 1023 x7/2= 21.077 x 1023=2.107 x 1024 molecules.
The chemical formula for ethane is C2H6 and there is 6.022 * 10^23 molecules in every mole. So to find your answer simply multiply 6.022 * 10^23 by 51.2 which gives: 3.083 X 10^25
The answer is 0,166.10e23 molecules.