A)Write the balanced chemical equation for this reaction.
2C2H2 + 5O2 ==> 4CO2 + 2H2O
B)Which is the limiting reactant?
I do these limiting reagent (LR) problems the long way.
mols C2H2 = g/molar mass = 13/26 = 0.5
mols CO2 produced if C2H2 is the LR.
0.5 x (4 mols CO2/2 mols C2H2) = 1
mols O2 = 13/32 = about 0.41
mols CO2 produced if O2 is the LR.
0.41 x (4 mols CO2/5 mols O2) = 0.32
The smaller number always wins since we can't produce more than the smallest amount. So O2 is the LR and C2H2 is the excess reagent (ER).
Express in chemical formula.
I have no idea what "express in chemical formula" means.
C)How many grams of C2H2 are present after the reaction is complete?
m=???g
mols O2 used = 0.41. mols C2H2 used = 0.41 x (2 mols C2H2/5 mols O2) = about 0.16. So C2H2 left over is 0.5 mols initially - 0.16 used = ?
Then grams = mols x molar mass
D)How many grams of O2 are present after the reaction is complete?
m=???g
You used all of the O2 since it was the LR.
E)How many grams of CO2 are present after the reaction is complete?
m=???g
F)How many grams of H2O are present after the reaction is complete?
m=???g
E and F are regular stoichiometry problems and are woked the same way as the first part. Be sure and check all the numbers since I've estimated and rounded here and there.Post your work if you get stuck.
When a mixture of 13.0 g of acetylene (C2H2) and 13.0 g of oxygen (O2) is ignited, the resultant combustion reaction produces CO2 and H2O.
A)Write the balanced chemical equation for this reaction.
B)Which is the limiting reactant?
Express in chemical formula.
C)How many grams of C2H2 are present after the reaction is complete?
m=???g
D)How many grams of O2 are present after the reaction is complete?
m=???g
E)How many grams of CO2 are present after the reaction is complete?
m=???g
F)How many grams of H2O are present after the reaction is complete?
m=???g
4 answers
@DrBob222
=.164 or .16
0.5 - .164= .336 or 0.5 - .16= .34
13g = mols X molar mass
what would the mols be? and
what would be the molar mass?
=.164 or .16
0.5 - .164= .336 or 0.5 - .16= .34
13g = mols X molar mass
what would the mols be? and
what would be the molar mass?
wrong
bad site. horrible
1 answer