posted by Eliazbeth on .
For the following reaction, 0.390 moles of nitrogen gas are mixed with 0.139 moles of oxygen gas.
nitrogen (g) + oxygen (g) nitrogen monoxide (g)
What is the FORMULA for the limiting reagent?
What is the maximum amount of nitrogen monoxide that can be formed? moles
N2 + O2 ==> 2NO
How many moles NO would form if we used all of the N2 and all of the O2 we needed. That is 0.590 x (2 moles NO/1 mole N2) = 1.18 moles NO formed.
How many moles NO would form if we used all of the O2 and all of the N2 we needed. That will be
0.139 x (2 moles NO/1 mol O2) = 0.278 moles NO.
Both answer CAN'T be right, of course. The correct answer in limiting reagent problems is ALWAYS the smaller number. Therefore, O2 is the limiting reagent and 0.278 mol NO will be formed. If you want grams, that is g = moles x molar mass.
When a sample of potassium nitrate is heated, oxygen gas is produced. This gas, collected in a 750 milliliter flask, has a pressure of 2.8 atmosphere and the temperature is recorded to be 53.6°C. How many moles of oxygen gas are produced in this reaction?
(R = 0.08205 L atm/K mol)
YOUR ANSWER IS .078