Answer: Theoretical yield is 313.6 g and the percent yield is, 91.8%
Explanation:
To calculate the moles :
According to stoichiometry :
1 mole of 
require 3 moles of 
Thus 2.8 moles of will require=
of
Thus is the limiting reagent as it limits the formation of product and is the excess reagent.
As 1 mole of give = 2 moles of 
Thus 2.8 moles of give =
of
Mass of 
Theoretical yield of liquid iron = 313.6 g
Experimental yield = 288 g
Now we have to calculate the percent yield
Therefore, the percent yield is, 91.8%
The % composition when 10g of magnesium combine with 4g of nitrogen is 71.43% magnesium and 28.57 % nitrogen
calculation
% composition = mass of an element / total mass x100
mass of magnesium = 10 g
mass of nitrogen = 4g
calculate the total mass used
= 10g of Magnesium + 4 g of nitrogen = 14 grams
% composition for magnesium is therefore = 10/14 x100 = 71.43 %
% composition for nitrogen is therefore = 4 /14 x100 = 28.57 %
Answer:
The reactants would appear at a higher energy state than the products.
Have a nice day!
Second one i think.......
