Asked by Saige
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?
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?
Answers
Answered by
Steve
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.
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.
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