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
wrong
@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?
bad site. horrible
A) To write the balanced chemical equation, we need to determine the products formed from the given reactants. Acetylene (C2H2) combines with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O).
The balanced chemical equation is:
C2H2 + 5O2 → 4CO2 + 2H2O
B) To determine the limiting reactant, we need to compare the number of moles of each reactant and see which one is consumed completely. We'll use the given masses to find the number of moles for each reactant.
The molar mass of acetylene (C2H2) is:
C2H2 = (2 * 12.01 g/mol) + (2 * 1.008 g/mol) = 26.038 g/mol
The molar mass of oxygen (O2) is:
O2 = 2 * 16.00 g/mol = 32.00 g/mol
To find the number of moles of acetylene:
moles of C2H2 = mass of C2H2 / molar mass of C2H2
moles of C2H2 = 13.0 g / 26.038 g/mol = 0.499 mol (approximately)
To find the number of moles of oxygen:
moles of O2 = mass of O2 / molar mass of O2
moles of O2 = 13.0 g / 32.00 g/mol = 0.406 mol (approximately)
Since the balanced equation has a 1:5 ratio between C2H2 and O2, we can see that we need 5 times more moles of O2 than C2H2 to react completely. Therefore, oxygen (O2) is the limiting reactant.
C) To find the mass of C2H2 remaining after the reaction is complete, we need to know the moles of C2H2 reacted. Since it is the limiting reactant, its moles will be completely consumed.
moles of C2H2 used = moles of O2 used
moles of C2H2 used = 0.406 mol (from previous calculations)
moles of C2H2 remaining = moles of C2H2 initially - moles of C2H2 used
moles of C2H2 remaining = 0.499 mol - 0.406 mol = 0.093 mol (approximately)
To find the mass of C2H2 remaining:
mass of C2H2 remaining = moles of C2H2 remaining * molar mass of C2H2
mass of C2H2 remaining = 0.093 mol * 26.038 g/mol = 2.419 g (approximately)
Therefore, after the reaction is complete, approximately 2.419 grams of C2H2 are present.
D) Since oxygen (O2) is the limiting reactant, it will be completely consumed in the reaction, leaving none remaining. Therefore, after the reaction is complete, there will be 0 grams of O2 present.
E) To find the mass of CO2 produced, we need to use the balanced equation.
From the balanced equation, we can see that for every 2 moles of C2H2 reacted, 4 moles of CO2 are produced.
moles of CO2 produced = moles of C2H2 used * (4 moles CO2 / 2 moles C2H2)
moles of CO2 produced = 0.406 mol * (4/2) = 0.812 mol
To find the mass of CO2 produced:
mass of CO2 = moles of CO2 * molar mass of CO2
mass of CO2 = 0.812 mol * (12.01 g/mol + 2 * 16.00 g/mol) = 36.320 g (approximately)
Therefore, after the reaction is complete, approximately 36.320 grams of CO2 are present.
F) Following the same logic as in part E, we find that for every 2 moles of C2H2 reacted, 2 moles of H2O are produced.
moles of H2O produced = moles of C2H2 used * (2 moles H2O / 2 moles C2H2)
moles of H2O produced = 0.406 mol * (2/2) = 0.406 mol
To find the mass of H2O produced:
mass of H2O = moles of H2O * molar mass of H2O
mass of H2O = 0.406 mol * 18.015 g/mol = 7.322 g (approximately)
Therefore, after the reaction is complete, approximately 7.322 grams of H2O are present.
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.