When Ca (s) and N2 (g) undergo a reaction, calcium nitride is?

Question

formed. What mass of clacium nitride can be formed when 56.6 gCa and 30.5 g of N2 is combined in a reaction vessel?

3Ca (s) + N2 (g) -------- Ca3N2(s)

Determine the limiting ractant?

Determine the excess reactant?

Determine the mass of Ca3N2 that can be formed?

Answer 1

3 moles Ca mass 3*40.08 = 120.24 g

1 mole N2 masses 2*14.01 = 28.02 g

56.6 / 120.24 ≈ 0.4707252

30.5 / 28.02 ≈ 1.088508

Determine the limiting reactant?  Ca

Determine the excess reactant? N2

Determine the mass of Ca3N2 that can be formed?

1 mol Ca3N2 masses 120.25 + 28.02 = 148.26 g

(0.4707252)(148.26 g) ≈ 69.78972 g ≈ 69.8 g