25g of NH3 is mixed with 4 moles of O2 in given reaction. 4NH3(g)+5O2(g)-->4NO(g)+6H2O(1) Which is limiting reactant?
Well, let's calculate the number of moles of NH3 and O2 we have.
Since the molar mass of NH3 is approximately 17 g/mol, we have:
25g NH3 * (1 mol NH3 / 17g NH3) = 1.47 moles of NH3
And we know that we have 4 moles of O2, as stated in the question.
Now let's look at the balanced equation.
The stoichiometric ratio between NH3 and O2 is 4:5. This means that for every 4 moles of NH3, we need 5 moles of O2.
Since we have 1.47 moles of NH3 and 4 moles of O2, it's clear that we have an excess of O2.
Therefore, the limiting reactant is NH3. It's like the O2 is saying, "Hey, NH3, stop hogging all the reaction and give me some space!"
Now, if we had a "Limiting Reactant of the Year" award, NH3 would definitely win because it's the one that determines the maximum amount of product that can be formed.
only answer
25g = 1.47 moles NH3
how much NH3 will 4 moles of O2 require?
Leencho
To determine the limiting reactant, we need to compare the number of moles of each reactant used in the reaction with their respective stoichiometric coefficients.
In the given reaction, the stoichiometric coefficients of NH3 and O2 are 4 and 5, respectively. We are given 4 moles of O2, so let's first calculate the number of moles of NH3.
The molar mass of NH3 (ammonia) is approximately 17 g/mol.
Number of moles of NH3 = Mass of NH3 / Molar mass of NH3
Number of moles of NH3 = 25 g / 17 g/mol
Number of moles of NH3 ≈ 1.47 moles
Now we can compare the moles of NH3 and O2. The mole ratio of NH3 to O2 from the balanced equation is 4:5.
Number of moles of NH3 : Number of moles of O2 = 1.47 : 4
Based on the mole ratio, it is clear that we have an excess of O2 since we have more moles of O2 available compared to NH3. Therefore, NH3 is the limiting reactant.
In summary, NH3 is the limiting reactant in the given reaction.