Calcium reacts with nitrogen. Calculate the mass of the product that is formed when 127.56 g of calcium reacts with 149.52 g of nitrogen. Identify your limiting reactant on this question.
3Ca + N2 --------> Ca3N2
moles of calcium present = 127.56/40.078
= 3.183 moles
moles of N present = 149.52/14.0067
= 10.675 moles
From the above equation, 3 moles of Ca are needed to completely react with 1 mole of N.
Ca is therefore used up during the reaction (less than one mole remains after 3 moles are used) while nitrogen has 9.675 moles left after 1 mole is used during the reaction.
Ca is the limiting reactant in the above reaction (if more product is to be formed, more Ca is to be added to the reactants)
mass of product = (40.078 * 3) + (14.0067*2)