March 5, 2015

Homework Help: CHEMISTRY...........

Posted by Anonymous on Friday, October 11, 2013 at 10:58pm.

1. If 29.0 L of methane, CH4, undergoes complete
combustion at 0.961 atm and 140C, 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
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 other words,there might be some follow-up questions.....

Answer this Question

First Name:
School Subject:

Related Questions

CHEMISTRY - 1. If 29.0 L of methane, CH4, undergoes complete combustion at 0.961...
Chemistry - Please help and explain each problem. PLEASE!!!! 1. If 29.0 L of ...
AP Chemistry - 33.0 L of methane (CH4) undergoes complete combustion at 0.961 ...
Chemistry - Natural gas is almost entirely methane, CH4. What volume of natural ...
Chemistry - a)What mass of water is produced from the complete combustion of 2....
chemistry - How many liters of water vapor can be produced if 8.9 liters of ...
skyline high school - Determine how many moles of water vapor will be produced ...
Chemistry - The complete combustion of 0.100 mol of CH4 in oxygen in a bomb ...
chemistry - Use Table 9.4 to calculate an approximate enthalpy or heat of ...