Areaction requires 22.4l of at stp. you have 32.0l of gas at 398k and 105.6 kpa. will you have enough gas to carry out the reaction? true or false

FALSE:

Explanation:

A reaction requires 22.4l of at STP. You have 32.0l of gas at 398k and 105.6 kpa.

1) STP stands for standard temperature and pressure.

2) Standard temperature is 0°C or 273.15 K

3) Standard pressure is 1 atm or 1013.25 kPa

4) use the ideal gas equation for both contidions

pV = n RT

=> n * R = pV /T

at STP n * R = 1031.25 kPa * 22.4 liter / 273.15 K = 84.5

at T = 398 K, p = 105.6 kPa, and V = 3.2.0 liter:

n * R = 105.6 kPa * 32.0 liter / 398 K = 8.49

Since R is a constant (the Universal Gases Constant), it is evident that the number of moles in the 32.0 liter of gas, at T = 398 K and P = 105.6 kPa is less than the number of moles of the 22.4 liter gas at STP.

There is not enough gas to carry out the reaction.