Answer:
12.5 g of Li are needed in order toproduce 0.60 moles of Li₃N
Explanation:
The reaction is:
6Li(s) + N₂(g) → 2Li₃N(s)
If nitrogen is in excess, the lithium is the limiting reactant.
Ratio is 2:6
2 moles of nitride were produced by 6 moles of Li
Then, 0.6 moles of nitride were produced by (0.6 .6)/ 2 = 1.8 moles of Li
Let's convert the moles to mass → 1.8 mol . 6.94 g/ 1mol = 12.5 g of Li
Answer:
Concentration of chloride ions = 0.584M
Explanation:
The step by step calculations is shown as attached below.
The volume of a sample of ammonia gas : 5.152 L
<h3>Further explanation</h3>
Given
0.23 moles of ammonia
Required
The volume of a sample
Solution
Assumed on STP
Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters / mol.
So for 0.23 moles :
= 0.23 x 22.4 L
= 5.152 L
Chemical Equation Balancer