How many grams of product are formed from 2.0 mol of n2 (g) and 8.0 mol of mg(s)? show all calculations leading to an answer. list 3 indicators of a chemical change.

Balanced chemical reaction happening here is:

3Mg(s) + N₂(g) → Mg₃N₂(s)        

 moles of product formed from each reactant:

2.0 mol of N2 (g) x 1 mol Mg₃N₂        = 2 mol Mg₃N₂

                                    1 mol N2

and

8.0 mol of Mg(s) x 1 mol Mg₃N₂         = 2.67 mol Mg₃N₂

                                 3 mol Mg

Since N2 is giving the least amount of product(Mg₃N₂) ie. 2 mol Mg₃N₂

N2 is the limiting reactant here and Mg is excess reactant.

Hence mole of product formed here is 2 mol Mg₃N₂    

molar mass of Mg₃N₂    

= 3 Mg + 2 N

= 101g/mol  

mass of product(Mg₃N₂) formed  

= moles x Molar mass

= 2 x 101

= 202g Mg₃N₂

202g of product are formed from 2.0 mol of N2(g) and 8.0 mol of Mg(s).

    The following are indicators of chemical changes:

Change in Temperature    

Change in Color

Formation of a Precipitate

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"