If 20.0 grams of propane are burned with 50.0 grams of oxygen:

Question

What mass of CO₂ is produced?
What mass of water is produced?
What mass of which reactant was excess?

Thank you so much.

Answer 1

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.

Answer 2

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

Answer 3

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
:)