To determine the limiting reactant, we need to compare the amount of N2 and O2 given in the question and see which one will be completely consumed first.
From the balanced equation N2 + O2 -> 2NO, we can see that one mole of N2 reacts with one mole of O2 to produce two moles of NO.
1. Calculate the moles of NO that can be produced from 2.25 mol N2:
2.25 mol N2 x (2 mol NO / 1 mol N2) = 4.50 mol NO
2. Calculate the moles of NO that can be produced from 2.00 mol O2:
2.00 mol O2 x (2 mol NO / 1 mol O2) = 4.00 mol NO
Since the amount of NO from N2 (4.50 mol) is greater than the amount from O2 (4.00 mol), O2 is the limiting reactant.
Now, we can calculate the number of moles and grams of NO formed from 2.00 mol O2:
Using the molar mass of NO (nitric oxide) which is 30.01 g/mol:
4.00 mol NO x 30.01 g/mol = 120.04 g NO
Therefore, 4.00 moles of O2 will produce 120.04 grams of NO.
Use the balanced equation, N2 + O2 2NO, to answer the following questions. Part: 0 / 3 Part 1 of 3 Determine the limiting reactant. Select the single best answer. 2.25 mol N2 2.00 mol O2 X How many moles of the product NO form? Be sure your answer has the correct number of significant figures. mol NO Calculate the number of grams of the product NO that form. Be sure your answer has the correct number of significant figures. NO
1 answer
1 answer