Answer:
Molarity = 0.7 M
Explanation:
Given data:
Volume of KCl = 20 mL ( 0.02 L)
Molarity = 3.5 M
Final volume = 100 mL (0.1 L)
Molarity in 100 mL = ?
Solution:
Molarity = number of moles of solute / volume in litter.
First of all we will determine the number of moles of KCl available.
Number of moles = molarity × volume in litter
Number of moles = 3.5 M × 0.02 L
Number of moles = 0.07 mol
Molarity in 100 mL.
Molarity = number of moles / volume in litter
Molarity = 0.07 mol /0.1 L
Molarity = 0.7 M
<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.
