The combustion of propane occurs when propane interacts with oxygen gas to produce water and carbon dioxide in the following reaction:
c3h8 (g) + 5o2 (g) → 3co2 (g) + 4h2o(l)
if 2 moles of propane c3h8 (g) react at 373 k in a volume of 2.5 l, what is the pressure in the cylinder from the c3h8 (g)?

Explanation:

1) Data:

a) n = 2 moles

b) T = 373 K

c) V = 2.5 liter

d) P = ?

2) Chemical principles and formula

You need to calculate the pressure of the propane gas in the mixture before reacting. So, you can apply the partial pressure principle which states that each gas exerts a pressure as if it occupies the entire volume.

Thus, you just have to use the ideal gas equation: PV = nRT

3) Solution:

P = 2 mol × 0.08206 atm-liter /K-mol × 373K / 2.5 liter = 24.5 atm

Since the number of moles are reported with one significant figure, you must round your answer to one significant figure, and that is 20 atm (20 is closer to 24.5 than to 30).

in scientific notation, the value of the  speed of light in a vacuum

  will be written as 2.995 * 10^8. when this value is represented in the form of

  a * 10^b, then, a = 2.995 and b = 8.

the value of the speed of light in a vacuum given above can also be rounded up to approximately 300000000. in this case, the scientific notation will be written as 3 *10^8.