In raction NH3 + 5 O2 --> 4 NO + 6 H20 . I 1 mol NH3 and 1 mol O2. what is the limiting reactant? ow Much NO will form?
To determine the limiting reactant and the amount of NO formed, we need to compare the stoichiometric ratios of NH3 and O2 to the desired product, NO.
First, let's calculate the number of moles of NO that can be formed from the given amounts of NH3 and O2.
From the balanced chemical equation: 1 mol NH3 reacts to produce 4 mol NO
Therefore, if we have 1 mol NH3, it will produce 4 mol NO.
Now, let's look at the stoichiometric ratio between O2 and NO.
From the balanced chemical equation: 5 mol O2 reacts to produce 4 mol NO
Therefore, if we have 1 mol O2, it will produce 4/5 mol NO.
To determine the limiting reactant:
1. Compare the moles of NO obtained from 1 mol NH3 to the moles of NO obtained from 1 mol O2.
- From 1 mol NH3, we can obtain 4 mol NO.
- From 1 mol O2, we can obtain (4/5) mol NO.
Since (4/5) is greater than 4, we can conclude that the O2 is the limiting reactant because it produces fewer moles of NO.
To calculate the amount of NO formed:
Since O2 is the limiting reactant and we have 1 mol of O2, we can determine how many moles of NO will form using the stoichiometric ratio:
From 1 mol O2, we can obtain (4/5) mol NO.
Therefore, if we have 1 mol O2, 4/5 mol NO will form.
In this case, as we have 1 mol O2, we can find that (4/5) moles of NO will be formed.