Chemistry Help needed ASAP

Question

The balanced equation is C3H8+ 5O2 --------> 3CO2+4H2O. Please show the work for A-D

A- Which one is the limiting reagent after 15.0 g of C3H8 reacts with19.5 g O2.

B- Determine the mass of excess reagent required for the above reaction

C- Determine the theoretical yield of H2O

D- Given the actual yield of H2O was 4.43 g, determine the percent yield.

Thanks so much!

Answer 1

A) First, need to convert the masses to moles. The reason for this is the stoichiometry of the equation is based on moles, not on mass.

To Calculate the molar masses:

Molar mass of C3H8 = (12.012*3) + (1.008*8) = 44.1 grams C3H8/mol

Molar mass of O2 = 16*2 = 32 grams O2/mol

To calculate how many moles of each reactant you have start with the masses given and use unit conversion to solve:

15 g C3H8 x 1mol/44.1 g C3H8 = 0.34 moles C3H8

19.5 g O2 x 1 mol/32 g O2 = 0.59 moles O2

Next take one of the known amount of moles of a reactant and compare using the stoichiometry of the reaction. In this example I use C3H8. Since it takes 5 moles of O2 to react with one mole of C3H8:

0.34 mol C3H8 x 5 mol O2/1 mol C3H8 = 1.7 mol O2

What this is saying is that if I have 0.34 mol of C3H8, then I would need 1.7 mol O2 to completely react the C3H8 I have. Since I have less moles of O2 than this, then my limiting reactant is O2.

B) Mass of excess reagent required for what? I am not sure the question is worded correctly

C) Assuming that I am using the numbers from part A)......here is how it would go:

Figuring out in A) that O2 is my limiting reagent, I need to find out how many moles of water I would produce. This is done my using the stoichiometry (relating one compound in the reaction to the other). We know it takes 5 moles of O2 to make 4 moles of water. Given this:

0.59 mol O2 x 4 mol H2O/5 mol O2 x 18 g H2O/1 mol H2O = 8.496 g H2O

D) Again assuming I am using Part C to answer the question, if you calculated in part C you have 8.496 g H2O, but you only made 4.43 g H2O in actuality, then the %yield is:

%yield = actual yield/theoretical yeild x 100%

%yield = 4.43 gH2O/8.496 g H2O x 100% = 52.1%

Hope this helps!