When ethane burns, it produces carbon dioxide and water: 2c2h6 (g) + 7o2 (g) > 4co2 (g) + 6h2) (i) how many molecules of water will be produced when 3 moles of ethane are burned

Hi!

To begin with, you need to understand that one mole of any molecule is equivalent to the Avogadro's constant, which is 6.022 x 10^(23).

To obtain the number of molecules for a particular number of moles of a molecule, you need to multiply the number of moles with the Avogadro's constant.

To calculate the number of molecules of water produced by the combustion of 3 moles of ethane, we need to first calculate the number of moles of water produced as a result of this reaction, and then multiply it with Avogadro's constant.

In the presence of abundant oxygen, ethane undergoes a complete combustion reaction (burns) to yield Carbon Dioxide and Water.

The balance equation for a combustion reaction in terms of whole numbers is for combustion of 2 moles of ethane:

2C2H6 (g) + 7O2 (g) --> 4CO2 (g) + 6H2O (I)

From this we can calculate the number of moles of water that will be produced as a result of the combustion of one mole of ethane, that is by dividing respective moles of each compound by 2 to give the equation:

1C2H6 (g) + 3.5O2 (g) --> 2CO2 (g) + 3H2O (l)

Now that we have obtained the balanced equation for the complete combustion of 1 mole of ethane, we can use this information to calculate the molecules of water produced as a result of combustion of 3 moles of ethane, that is by multiplying each number of mole by 3.

3C2H6 (g) + 10.5O2 (g) --> 6CO2 (g) + 9H2O (l)

It is important to note here that we have calculated the number of moles of water that have been produced, which is 9, and we need to multiply this  with the Avogadro's constant:

9 x (6.022 x 10^(23)) = 5.4198 x 10^(24)

~ 5.42 x 10^(24) molecules of water

5.42 x 10^(24)

Explanation:

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g

points

explanation: