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.
How many grams of C2H2 are present after the reaction is complete?
How many grams of CO2 are present after the reaction is complete?
How many grams of H2O are present after the reaction is complete?
@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?
Note that I said "about" 0.16. Since I rounded and used 0.41, which is also an about, I rounded to 2 decimals. You should calculate to 3 places if you wish but if so then redo the 0.41 also to see that that numbers is. I think it's 0.406 to three places.
I also calculated the molar mass of C2H2 (28) and O2 (32).
To determine the grams of C2H2, CO2, and H2O present after the reaction is complete, we need to follow a series of steps. Here's how you can calculate each quantity:
Step 1: Determine the balanced chemical equation for the combustion reaction:
C2H2 + O2 --> CO2 + H2O
Step 2: Calculate the moles of C2H2 and O2 using the given masses:
Molar mass of C2H2 = 2(12.01 g/mol) + 2(1.01 g/mol) = 26.04 g/mol
Moles of C2H2 = 13.0 g / 26.04 g/mol = 0.499 mol
Molar mass of O2 = 2(16.00 g/mol) = 32.00 g/mol
Moles of O2 = 13.0 g / 32.00 g/mol = 0.406 mol
Step 3: Identify the limiting reactant:
The limiting reactant is the one that is completely consumed in the reaction and limits the amount of product formed. To determine the limiting reactant, we compare the mole ratio of the reactants from the balanced equation. The ratio of C2H2 to O2 is 1:1, which means that for every 1 mol of C2H2, we need 1 mol of O2. However, in this case, we have more moles of C2H2 (0.499 mol) than O2 (0.406 mol). Therefore, O2 is the limiting reactant.
Step 4: Calculate the moles of CO2 and H2O produced based on the stoichiometry of the balanced equation:
From the balanced equation, we see that the mole ratio of C2H2 to CO2 is 1:1, and the mole ratio of C2H2 to H2O is also 1:1. This means that for every 1 mol of C2H2, we get 1 mol of CO2 and 1 mol of H2O.
So, the moles of CO2 and H2O produced are both 0.406 mol.
Step 5: Calculate the masses of C2H2, CO2, and H2O:
Mass of C2H2 = Moles of C2H2 x Molar mass of C2H2
Mass of C2H2 = 0.499 mol x 26.04 g/mol = 12.99 g
Mass of CO2 = Moles of CO2 x Molar mass of CO2
Mass of CO2 = 0.406 mol x 44.01 g/mol = 17.86 g
Mass of H2O = Moles of H2O x Molar mass of H2O
Mass of H2O = 0.406 mol x 18.02 g/mol = 7.33 g
Therefore, after the reaction is complete:
- The grams of C2H2 present are 12.99 g
- The grams of CO2 present are 17.86 g
- The grams of H2O present are 7.33 g.
I worked this part of a multipart question for K.G. Here is the link. Post your work if you have questions.
https://www.jiskha.com/questions/1816833/when-a-mixture-of-13-0-g-of-acetylene-c2h2-and-13-0-g-of-oxygen-o2-is-ignited-the