Help with chemistry??? WILL FAN AND MEDAL ?
Okay, well I'm so confused with stoichiometry, could you please explain how in the problem they got the numbers 22.4 L O2 and the 2 mol C and 1 mol O2 as well as the 12.01, where did they get these numbers from? I'm confused. How many grams of Carbon are needed to react completely with 32.8 liters of O2 gas at STP? C(s)+O2(g)-->2CO(g) 32.8 LO2 x (1 mol O2)/(22.4 L O2) x (2 mol C)/(1 mol O2) x (12.01 g C)/ (1 mol C)=35.2 g C
At STP volume of 1 mole of O2 = 22.4L using this they have found out 32.8L is how many moles... okay?
So it is : 32.8 LO2 x (1 mol O2)/(22.4 L O2) =1.46 moles of O2 . Now, we know that 1 mole of O2 weighs 32g then how much will 1.46 moles weigh?
NOTE: the equation you have written isn't balanced properly... pls. write the correct version
@Jen543 ?
carbon combines with oxygen to form carbon monoxide 2C + O2 --. 2CO this is a balanced equation because there are 2 atoms of carbon and 2 atoms of oxygen on left side and there are the same numebr of C's and O's on the right side.
if we check the periodic table we see that one mole of carbon weighs 12 grams and we also know that one mole of the gas oxygen occupies 22.4 liters at Standard Temp and pressure. (this is the volume of 1 mole of every gas).
understood so far?
So sorry for the late reply, yes thank you I appreciate it so much
Is this how you do every stoichiometry problem?
I'm still extremely confused with chem
you need to get the correct balanced equation first. 2C + O2 = 2CO 2 moles of carbon react with 22.4 litres of oxygen now 1 mole of carbon = 12 grams so 2*12 = 24 grams C reacts with 22.4 litres oxygen OK/
Thank you!
Ok so you need to find how much carbon will react with 32.8 liters of oxygen as we have seen already 24 g carbon react with 22.4 liters O2 trun this around:- 22.4 liters O2 react with 24 g Carbon by proportion 32.8 liters react with 32.8 * 24 --------- grams of carbon 22.4
do that calculation and you'll get the answer
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