general chemistry question

[Ferdowsi]

Advertisement - Members don't see this ad

If 88 g of C​

​

3H8 and 160 g of O2 are allowed to react
maximally to form CO2 and H2O, how many grams of​

CO2 will be formed?

so i found out that oxygen is limiting reagent. and then setting up:

160 gram of oxygen * 1/160 * 3/5*44/1: but this is not the right answer.. i don't get it

 

[Albuterol]

C3H8 + 5O2 ---> 3CO2 + 4H2O

Moles of CO2 from C3H8
88g x (1mol/44g) x (3mol CO2/1mol C3H8) = 6 moles CO2

Moles of CO2 from O2
160g x (1mol O2/32g) x (3mol CO2/5 mol O2)= 3 moles of CO2

Since O2 is LR
3 moles of CO2 x (44g CO2/1mol CO2) = 132g CO2

 

[Ferdowsi]

C3H8 + 5O2 ---> 3CO2 + 4H2O

Moles of CO2 from C3H8
88g x (1mol/44g) x (3mol CO2/1mol C3H8) = 6 moles CO2

Moles of CO2 from O2
160g x (1mol O2/32g) x (3mol CO2/5 mol O2)= 3 moles of CO2

Since O2 is LR
3 moles of CO2 x (44g CO2/1mol CO2) = 132g CO2

I don't get it🙁...its so frustrating...i get the limiting reagent but don't you need to start off with 160 gram of oxygen? ooooooooh fffk

 

[keliao]

88g x (1mol/ 44g C3H8) = 2 moles CO2 limiting reagent
160g (1mol/32g O2) = 5 moles O2

so (88g/44gC3H8)* (3 mol CO2/1 mol C3H8) * (44g CO2/1 mol CO2)= 264g CO2

is this correct?

 

[Albuterol]

Ok. you start off with 88 grams of C3H8 and 160grams of O2. Find the limiting reagent.
We see that C3H8 produces 6 Moles of CO2 and O2 produces 3 Moles of CO2. You realize that 3 moles of CO2 is the MOST we can produce so therefore 160 grams of O2 is the limiting reagent. So now that you have already found the #moles of CO2 produced from O2 all u need to do is multiply through with the g/mol of CO2. 3 moles CO2 x (44g/mol CO2) = 132g

160 gram of oxygen * 1/160 * 3/5*44/1: but this is not the right answer.. i don't get it
1/160 is NOT the g/mol of O2 ...g/mol of O2 is 32g/mol. this is where ur mistake is. you are using the wrong g/mol ratio.

 

[Albuterol]

88g x (1mol/ 44g C3H8) = 2 moles CO2 limiting reagent
160g (1mol/32g O2) = 5 moles O2

so (88g/44gC3H8)* (3 mol CO2/1 mol C3H8) * (44g CO2/1 mol CO2)= 264g CO2

is this correct?

this is not correct...u forgot to multiply with the CO2/C3H8 mole ratio. so really u have just found the # of moles of C2H8 and not # moles of CO2 produced when C3H8 is reacted.

 

[keliao]

to find te limiting reagent u dont need to multiply it by (3mol CO2/1mol C3H8)
you multiply it after you find the least number of moles