Answer:

Explanation:
We are given the mass of two reactants, so this is a limiting reactant problem.
We know that we will need mases, moles, and molar masses, so, let's assemble all the data in one place, with molar masses above the formulas and masses below them.
M_r: 17.03 32.00 18.02
4NH₃ + 5O₂ ⟶ 4NO + 6H₂O
m/g: 70.1 70.1
Step 1. Calculate the moles of each reactant

Step 2. Identify the limiting reactant
Calculate the moles of H₂O we can obtain from each reactant.
From NH₃:
The molar ratio of H₂O:NH₃ is 6:4.

From O₂:
The molar ratio of H₂O:O₂ is 6:5.

O₂ is the limiting reactant because it gives the smaller amount of H₂O.
Step 3. Calculate the theoretical yield.

Answer:
A. elements
<em>I</em><em> hope</em><em> it's</em><em> helps</em><em> you</em>
The answer to this would be false.
We are given with the total mass of fertilizer which is 14.1 oz. This is equivalent to 399.73 grams. 15% of the total is the amount of nitrogen. Thus, the nitrogen amount is 59.96 grams.