Solution Found!
When 35.6 L of ammonia and 40.5 L of oxygen gas at STP
Chapter 6, Problem 46P(choose chapter or problem)
When \(35.6 L\) of ammonia and \(40.5 L\) of oxygen gas at STP burn, nitrogen monoxide and water are produced. After the products return to STP, how many grams of nitrogen monoxide are present?
\(\mathrm{NH}_{3}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{NO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) [unbalanced]
Equation Transcription:
Text Transcription:
35.6 L
40.5 L
NH_3(g) + O_2(g) rightarrow NO(g)+H_2O(l)
Questions & Answers
QUESTION:
When \(35.6 L\) of ammonia and \(40.5 L\) of oxygen gas at STP burn, nitrogen monoxide and water are produced. After the products return to STP, how many grams of nitrogen monoxide are present?
\(\mathrm{NH}_{3}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{NO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) [unbalanced]
Equation Transcription:
Text Transcription:
35.6 L
40.5 L
NH_3(g) + O_2(g) rightarrow NO(g)+H_2O(l)
ANSWER:
Solution 46P
Here we have to find out mass of nitrogen monoxide(NO) has form in grams .
Step 1
Given:
Volume of NH3 = 35.6 L
Volume of oxygen gas = 40.5L
The balanced chemical equation for the reaction between ammonia and oxygen is given below,
4NH3(g) +5O2(g) → 4NO(g) +6H2O (l)
In order to find out the mass of NO, 1st we have to find out the limiting reactant, then by using the limiting reactant we can calculate the mass of NO formed.
NO from NH3:-
In the above reaction it has been found that 4 mole of NO has formed from 1 mole of NH3.
1st we will convert of NH3 into mole as follows,
It is known that 1 mole of NH3 = 24L
Thus 35.6 L of NH3 will be equal to = 35.6 L
= 1.48 mole
Thus the mole of NO formed can be calculated as,
Given mole of NH3
= 1.48 mole of NH3
= 1.48 mol NO
NO from O2:-
From the above balanced equation it has been found that 4 mole of NO has formed from 5 mole of O2.
1st we will convert of oxygen into mole as follows,
It is known that 1 mole of hydrogen = 24L
Thus 40.5 L of oxygen will be equal to = 40.5 L
= 1.68 mole
Thus the mole of NO formed can be calculated as,
Mole of O2
= 1.68 mole of O2
= 1.34 mol of NO
The mole of NO formed from O2 is less hence it is the limiting reactant.