1. If 29.0 L of methane, CH4, undergoes complete
combustion at 0.961 atm and 140°C, how many liters
of each product would be present at the same
temperature and pressure?
My answer:
(0.961 atm)(29.0 L) (413k) all divided by (413)(0.961 atm)=29.0 L CO2
I really don't understand what you've done here but 29.0L is the correct answer. The easy way to work this problem is to note that if everything is in the gaseous state then you need not convert everything to mols, do the mols conversion, then back to L. Instead you can use L as mols; therefore, 29.0L CH4 produces 1 mol CO2 for each mol CH4 and that is 29.0L CO2.
For H2O you have done it right; i.e., 1 mol CH4 produces 2 mols H2O; therefore, 29.0L x 2 = 58.0 L H2O (in the gaseous state of course).
29.0 L CO2 x 2LH2O/1LCO2=58.0 L H2O
1. If 29.0 L of methane, CH4, undergoes complete
combustion at 0.961 atm and 140°C, how many liters
of each product would be present at the same
temperature and pressure?
My answer:
(0.961 atm)(29.0 L) (413k) all divided by (413)(0.961 atm)=29.0 L CO2
29.0 L CO2 x 2LH2O/1LCO2=58.0 L H2O
2. If air is 20.9% oxygen by volume,
a. how many liters of air are needed for complete
combustion of 25.0 L of octane vapor, C8H18?
b. what volume of each product is produced?
****I do not know how to do a, but I got 200L CO2 and 225 L H2O vapor for part b.
3.Methanol, CH3OH, is made by causing carbon
monoxide and hydrogen gases to react at high
temperature and pressure. If 4.50 × 102 mL CO and
8.25 × 102 mL H2 are mixed,
a. which reactant is present in excess?
b. how much of that reactant remains after the
reaction?
c. what volume of CH3OH is produced, assuming the
same pressure?
*****a) the reactant in excess is CO because it produces 14 g CH3OH and H2 produces 13.2 g CH3OH...14>13.2
I need help on b) and c)...
I might be asking you some questions after you helped me, so keep track of my post....in other words,there might be some follow-up questions.....
4 answers
2. If air is 20.9% oxygen by volume,
a. how many liters of air are needed for complete
combustion of 25.0 L of octane vapor, C8H18?
b. what volume of each product is produced?
****I do not know how to do a, but I got 200L CO2 and 225 L H2O vapor for part b.
You do this the same way, up to a point, that you solve for CO2 and H2O (the 200L and 225L values are correct).
25L C8H18 x (25 mol O2/2 mol C8H18) = 312.5 L O2 (if it were pure O2).
Then 20.9% of what number = 312.5
0.209x = 312.5
x = about 1495L at 20.9% purity. Watch the number of significant figures; you're not allowed 4 places.
a. how many liters of air are needed for complete
combustion of 25.0 L of octane vapor, C8H18?
b. what volume of each product is produced?
****I do not know how to do a, but I got 200L CO2 and 225 L H2O vapor for part b.
You do this the same way, up to a point, that you solve for CO2 and H2O (the 200L and 225L values are correct).
25L C8H18 x (25 mol O2/2 mol C8H18) = 312.5 L O2 (if it were pure O2).
Then 20.9% of what number = 312.5
0.209x = 312.5
x = about 1495L at 20.9% purity. Watch the number of significant figures; you're not allowed 4 places.
3.Methanol, CH3OH, is made by causing carbon
monoxide and hydrogen gases to react at high
temperature and pressure. If 4.50 × 102 mL CO and
8.25 × 102 mL H2 are mixed,
a. which reactant is present in excess?
b. how much of that reactant remains after the
reaction?
c. what volume of CH3OH is produced, assuming the
same pressure?
*****a) the reactant in excess is CO because it produces 14 g CH3OH and H2 produces 13.2 g CH3OH...14>13.2
I need help on b) and c)...
I'm assuming high T and P means everything is gaseous. You're right with a that H is the limiting reagent (LR) and CO is the excess reagent (ER). You converted to grams but you can keep everything in the gaseous state and stick with mL.
b.
If H2 is the LR, then how much CO will be used for complete reaction? That is
825 mL H2 x (1 mol CO/2 mol H2) = 412.5 mL CO used.
We had 450 initially so 450-412.5 = ? mL left unreacted.
c.
Assuming high T and P (so CH3OH is a gas) the reaction will produce 825 mL H2 x (1 mol CH3OH/2 mol H2) = 825 x 1/2 = ? mL CH3OH VAPOR.
Again, watch the number of s.f. You're allowed 3. from 8.25E2 and 4.50E2.
monoxide and hydrogen gases to react at high
temperature and pressure. If 4.50 × 102 mL CO and
8.25 × 102 mL H2 are mixed,
a. which reactant is present in excess?
b. how much of that reactant remains after the
reaction?
c. what volume of CH3OH is produced, assuming the
same pressure?
*****a) the reactant in excess is CO because it produces 14 g CH3OH and H2 produces 13.2 g CH3OH...14>13.2
I need help on b) and c)...
I'm assuming high T and P means everything is gaseous. You're right with a that H is the limiting reagent (LR) and CO is the excess reagent (ER). You converted to grams but you can keep everything in the gaseous state and stick with mL.
b.
If H2 is the LR, then how much CO will be used for complete reaction? That is
825 mL H2 x (1 mol CO/2 mol H2) = 412.5 mL CO used.
We had 450 initially so 450-412.5 = ? mL left unreacted.
c.
Assuming high T and P (so CH3OH is a gas) the reaction will produce 825 mL H2 x (1 mol CH3OH/2 mol H2) = 825 x 1/2 = ? mL CH3OH VAPOR.
Again, watch the number of s.f. You're allowed 3. from 8.25E2 and 4.50E2.
If you have follow up questions please
a. post to my attention AND THIS IS IMPORTANT
b. Copy the problem and answers followed by your follow up question; otherwise, this post can get lost in the shuffle and I'll not be able to find it.
a. post to my attention AND THIS IS IMPORTANT
b. Copy the problem and answers followed by your follow up question; otherwise, this post can get lost in the shuffle and I'll not be able to find it.
1 answer