Question

12.If each symbol represents 1 mole of Mg (s) and 1 mole of N2 (g), and the the reactants are mixed in the proportions shown, what is the limiting reactant and how many moles of excess reactant remain after the reaction is complete? a) N2 (g), 2 mole Mg (s) b) N2 (g), 3 mole Mg (s) c) Mg (s), 1 mole N2 (g) d) Mg (s), 2 mole N2 (g) e) Mg (s), 0.5 mole N2 (g)

Answer 1

the reation must be

3Mg + N2 = Mg3N2

then, ratio is 3:1

ratio = Mg/N2 = 3/1 = 3

a)

1 mol of N2, 2 mol of Mg

ratio

Mg/N = 2/1 = 0.5

since 0.5 < 3, then Mg is limiting

1-2/3 = 0.333 mol of Mg left

b)

3/1 = 3

no limiting reactant since ratio = 3, similar as 3

0 mol left,

c)

1:1

clearly, there is not enough Mg to react

1-1/3 = 0.6667 mol of Mg left

d)

1 mol of }MG = 2 mol of N2

Mg is limiting, since 1/2 = 0.5

2-1/3 = 1.666 mol of N2 excess

e)

1 mol of MG = 0.5 mol of N2

1/0.5 = 2

since 2 < 3, there is not enough MG

excess = 0.5 - 1/3 = 0.1667 mol of MG