In one experiment, 0.893 mol of NO is mixed with 0.519 mol of O2. Determine which of the two reactants is the limiting reactant. Calculate also the number of moles of NO2 produced.

NO is the limiting reactant.

0.893 moles of NO2 will be produced

Explanation:

Step 1: Data given

Number of moles NO = 0.893 moles

Number of moles O2 = 0.519 moles

Step 2: The balanced equation

2NO + O2 → 2NO2

Step 3: Calculate the limiting reactant

For 2 moles NO we need 1 mol O2 to produce 2 moles NO2

NO is the limiting reactant. It will completely be consumed. (0.893 moles)

O2 is in excess. There will react 0.893/2 = 0.4465 moles.

There will remain 0.519 - 0.4465 = 0.0725 moles O2

Step 4: Calculate moles NO2

For 2 moles NO we need 1 mol O2 to produce 2 moles NO2

For 0.893 moles NO we'll have 0.893 moles NO2

NO is the limiting reactant.

0.893 moles of NO2 will be produced

the answer is 303.15 or 303

the maximum mass of nh3 is nh3 in grams