Answer:
Question 2: Na3PO4, KOH; Question 3: Na3PO4, KOH
Explanation:
Question 2
The reactants in a chemical equation are the species on the left side of the reaction arrow.
Thus the reactants are Na3PO4, KOH (sodium phosphate and potassium hydroxide).
Question 3.
The products in a chemical equation are the species on the right side of the reaction arrow.
Thus the products are NaOH, K3PO4 (sodium hydroxide and potassium phosphate).
You have to switch the elements based in if they're metals/nonmetals. so nonmetals switch with each other and metals switch with each other.
AB + CD ---> AC + BD
Answer:- There are 
moles.
Solution:- It is a unit conversion problem where we are asked to convert mg of aspartame to moles. Aspartame is 
and it's molar mass is 294.31 grams per mole.
mg are converted to grams and then the grams are converted to moles as:
= moles of aspartame
So, there would be moles of aspartame in 1.00 mg of it.
Number one victory royal yaaa
<h3>Answer:</h3>
Limiting reactant is Lithium
<h3>
Explanation:</h3>
<u>We are given;</u>
- Mass of Lithium as 1.50 g
- Mass of nitrogen is 1.50 g
We are required to determine the rate limiting reagent.
- First, we write the balanced equation for the reaction
6Li(s) + N₂(g) → 2Li₃N
From the equation, 6 moles of Lithium reacts with 1 mole of nitrogen.
- Second, we determine moles of Lithium and nitrogen given.
Moles = Mass ÷ Molar mass
Moles of Lithium
Molar mass of Li = 6.941 g/mol
Moles of Li = 1.50 g ÷ 6.941 g/mol
= 0.216 moles
Moles of nitrogen gas
Molar mass of Nitrogen gas is 28.0 g/mol
Moles of nitrogen gas = 1.50 g ÷ 28.0 g/mol
= 0.054 moles
- According to the equation, 6 moles of Lithium reacts with 1 mole of nitrogen.
- Therefore, 0.216 moles of lithium will require 0.036 moles (0.216 moles ÷6) of nitrogen gas.
- On the other hand, 0.054 moles of nitrogen, would require 0.324 moles of Lithium.
Thus, Lithium is the limiting reagent while nitrogen is in excess.
