Help me in chemistry MOLES ?

In the formation reaction :


2C+3H2----%26gt;C2H6





Calculate the number of moles of C2H6 formed when


a) 2.0 moles of C are reacted with 5.0 moles of H2


b) 6.0 moles of H2 are reacted with 4.0 moles C


c) 9.0 moles of H2 are reacted with 5.0 moles of C


d) 0.0812 moles of C reacted with 0.125 moles of H2

Help me in chemistry MOLES ?
look for the limiting component in each case:


1) only 1 mol of C2H6 is form (2 moles of C can only make 1 moles of C2H6)


2) 2 mol (6 mols H2 = 3 mol C2H6)


3) 2.5 mol


4) 1.9488 mol
Reply:Think of moles as the number in front of reagents and products. in a) C is the limiting reagent. Therefore only


one mole of product as per equationb) same ratio of moles


of reagents therefore proportional moles of product.


c) finally a challenge. Which is limiting: 3:2 or 9:5: Not enough\


carbon. Use up all the carbon.and 5*3/2=7.5 of hydrogen


for 2.5 moles of ethane. Last one is yours
Reply:These are limiting reagents problems. Basically one component is going to be used up before the other, and you have to find out which. To do this, you need to use your balanced equation, this one says that for every 2 moles of C you will need 3 moles of H2 to give you 1 mole of C2H6.





a) 2.0 moles of C reacting with 5.0 moles of H2.


Find the limiting reagent first





2.0 mol C x 3 mol H2 / 2mol C = 3 mol H2


This says that for 2 moles of C you need 2 moles of H2, since you have 5.0 moles you got plently, and the 2.0 mole C is you LR.





2.0 mol C x 1 mol C2H6/ 2 mol C = 1.0 mol C2H6





b) 6.0 moles H2 react w/ 4.0 mol C





Find the LR


6.0 mol H2 x 2 mol C / 3 mol H2 = 4 mol C





6 mol H2 needs EXACTLY 4 mol C, which you have


6.0 mol H2 x 1 mol C2H6 / 3 mol H2 = 2.0 mol C2H6





c) 9.0 mol H2 w/ 5.0 mol C


Find the LR


9.0 mol H2 x 2 mol C / 3 mol H2 = 6 mol C


9.0 moles H2 needs 6 mols of C. You have only 5 mol C so 5 mol C is the LR





5.0 mol C x 1 mol C2H6 / 2 mol C = 2.5 mol C2H6





d) 0.0812 mol C react w/ .125 mol H2


Find the LR first


.0812 mol C x 3 mol H2 / 2 mol C = .1218 mol H2


.1218 mol H2 %26lt; .126 mol H2, so the .0812 mol C is the LR





.0812 mol C x 1 mol C2H6 / 2 mol C = 0.0406 mol C2H6