Question

in a combination reaction if 1.30688 moles of magnesium is heated with .2977 moles of nitrogen to form magnesium nitride, how many grams of magnesium nitride will be produced

Answer 1

3Mg + N2 -- > Mg3N2
according to reaction
1 mole of N2 required 3 mole of Mg
0.2977 mole of N2 required (3)*0.2977 mole of Mg
0.2977 mole of N2 required 0.8931 mole of Mg
but we have 1.30688 mole of Mg
so, Mg are in excess
and N2 are limiting reagent

again, according to reaction
1 mole of N2 give 1 mole of Mg3N2
0.2977 mole of N2 give 0.2977 mole of Mg3N2
number of mole of Mg3N2 formed = 0.2977 mole

molar mass of Mg3N2 = 100.9 g/mol
mass of Mg3N2 formed= (number of mole of Mg3N2)*(molar mass of Mg3N2)
= (0.2977*100.9) g
= 30.04 g

Answer : 30.04 g