Answer:
3.62moles
Explanation:
First let us generate a balanced equation for the reaction
2Na + Cl2 —> 2NaCl
Molar Mass of Na = 23g/mol
Mass of Na from the balanced equation = 2 x 23 = 46g
Molar Mass of NaCl = 23 + 35.5 = 58.5g/mol
Mass of NaCl from the balanced equation = 2 x 58.5 = 117g
From the question,
46g of Na produced 117g of NaCl.
Therefore, 83g of Na will produce = (83 x 117)/46 = 211.11g of NaCl
Converting this mass (211.11g of NaCl) to mole, we obtain:
n = Mass /Molar Mass
n = 211.11/ 58.5
3.62moles
Answer:
Explanation:
Just saw your request regarding answering this so here it is:
All of them belong of Group 1 in periodic table and thus are highly reactive! Pattern of reactivity for Group 1 (Alkali metals) increases as you move down the group as their radius keeps increasing and thus electrons can be easily lost. Thus, to ID the lumps, Sheena should look at their reactivity and she should get the following trend:
Most reactive: Potassium (K)
Intermediate: Sodium (Na)
Least reactive: Lithium (Li)
Hope it helps!
Answer:
Electrons orbit the nucleus.
Answer:
The percent yield of chloro-ethane in the reaction is 82.98%.
Explanation:
Moles of ethane = 
Moles of chlorine gases =
As we can see that 1 mol of ethane react with 1 mole of chlorine gas.the 10 moles will require 10 mole of chlorine gas, but only 9.1549 moles of chlorine gas is present.
This means that chlorine gas is in limiting amount and amount of formation of chloro-ethane will depend upon amount of chlorine gas.
According to reaction , 1 mol of chloro ethane gives 1 mol of chloro-ethane.
Then 9.1549 moles of chlorien gas will give:
of chloro-ethane
Mass of 9.1549 moles of chloro-ethane:
9.1549 mol × 64.5 g/mol = 590.4910 g
Theoretical yield of chloro-ethane: 590.4910 g
Given experimental yield of chloro-ethane: 490.0 g
The percent yield of chloro-ethane in the reaction is 82.98%.
