2.22 g of Mg is 2.22/24.31 = 0.09 mole
3.75g of N is 3.75/14.01 = 0.27 mole
Now, what compound is formed? The ratio of elements will affect how much reagent is used.
If we assume the simplest Mg3N2 then 3 moles of Mg are used for each 2 moles of N.
So, .09 mole of Mg will react with 2/3 * .09 = .06 mole of N.
You can take it from here, I guess.
In a combination reaction, 2.22 g of Magnesium is heated with 3.75 g of Nitrogen
a) which reactant is present in excess?
b) how many moles of product are formed?
c) after reaction, how many grams of each reactant and product are present?
1 answer