Answer:
?
Explanation:
What are the statements? You've given the passage but not the statements
<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.
First and foremost, they are completely different substances with each exhibiting unique properties. Both have different atoms involved on their structures which is the cause of the differing properties.
Answer:
Away from the central sulfur atom.
Explanation:
Answer:
141g of CCl₄
Explanation:
First, we have to write the balanced equation.
CCl₄(g) + 2 HF(g) ⇄ CF₂Cl₂(g) + 2 HCl(g)
We can calculate how many moles of CF₂Cl₂ using the ideal gas equation.
V = 14.9 dm³ = 14.9 L
T = 21°C + 273.15 = 294.15 K
P = 1.48 atm
R = 0.08206 atm.L/mol.K

We can use proportions to find the mass of CCl₄ required to obtain 0.914 moles of CF₂Cl₂. According to the balanced equation, 1 mol of CF₂Cl₂ is produced when 1 mol of CCl₄ reacts. And the molar mass of CCl₄ is 154 g/mol.
