<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.
At STP, 1 mol = 22.4
1 mol = 6.022 X 10^ 23
so your answer is right it is B 6.022 X 10^ 23
I don’t know but I think it would be products... that’s the best I can give. I’ll look more into it
I believe your answer should be C because the heat would move from the soup to the spoon.
