If 20.0 grams of propane are burned with 50.0 grams of oxygen:
What mass of CO₂ is produced?
What mass of water is produced?
What mass of which reactant was excess?
Thank you so much.
I work these by solving two simple stoichiometry problems. Here is a link to how to do those but I think I've shown you before.
http://www.jiskha.com/science/chemistry/stoichiometry.html
You will obtain two different answers but the correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent. After you do the first set (say for CO2) and identified the limiting reagent it is not necessary to do the second set after that. Post your work if you get stuck.
change both masses to moles.
write the balancedequation.
C3H8+5O2>>3CO2 + 4H2O
Now,it find the limiting reactant.
it takes 5 moles of O2 for each mole of propane.
Moles Propane= 20/44=.455
moles O2=50/32=1.56
so you have mole ratio of 1.56/.455=3.43, so you do not have enough oxygen, so oxygen is the limiting reactant.
You react then 3.43moles O2, you get 3/5*3.43 moles CO2, and 4/5*3.43 moles H2O convert those to grams
stoichiometry song:
*to the tune of hark the herald angels sing* (we learned this last year at christmas time:P)
Stoichiometry is easy with the five step recipe
balance chemical equation
I.D. unknown and given
Change the given into moooooles
use the equation ratioooo
put the unknown on the top
bottom the given but don't stop
Change unknown moles into the
measure the question asked of thee
:)
To determine the mass of CO₂ produced, the mass of water produced, and which reactant is in excess, we need to use stoichiometry and the balanced chemical equation for the combustion of propane.
The balanced chemical equation for the combustion of propane is:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
First, we need to determine which reactant is limiting and which is in excess. To do this, we have to compare the number of moles of each reactant to their respective stoichiometric coefficients in the balanced equation.
1. Calculate the number of moles for each reactant:
Number of moles of propane (C₃H₈) = mass of propane (20.0 g) / molar mass of propane
Number of moles of oxygen (O₂) = mass of oxygen (50.0 g) / molar mass of oxygen
The molar mass of propane (C₃H₈) is calculated as:
(3 * atomic mass of carbon) + (8 * atomic mass of hydrogen)
The molar mass of oxygen (O₂) is simply the atomic mass of oxygen.
2. Compare the ratios of moles of each reactant to their stoichiometric coefficients:
For propane, divide the number of moles by the coefficient of propane (C₃H₈) in the balanced equation (1) to get the mole ratio of propane to CO₂ and H₂O.
For oxygen, divide the number of moles by the coefficient of oxygen (O₂) in the balanced equation (5) to get the mole ratio of oxygen to CO₂ and H₂O.
The reactant that produces fewer moles of products will be the limiting reactant.
3. Determine the limiting reactant:
Compare the mole ratios of moles propane to CO₂ and moles oxygen to CO₂ to see which one is smaller. The reactant with the smaller mole ratio is the limiting reactant.
4. Calculate the moles of CO₂ produced:
Use the mole ratio from the balanced equation. In this case, the mole ratio of propane to CO₂ is 1:3.
5. Calculate the mass of CO₂ produced:
Multiply the moles of CO₂ produced by the molar mass of CO₂ to get the mass of CO₂.
6. Calculate the mass of water produced:
Use the mole ratio from the balanced equation. In this case, the mole ratio of propane to water is 1:4.
7. Determine the mass of excess reactant:
Subtract the mass of the limiting reactant from the initial mass of the reactant to determine the mass of the excess reactant.
By following these steps, you can calculate the mass of CO₂ produced, the mass of water produced, and identify which reactant was in excess in the given combustion reaction.