Question

Write and balance the combination reaction for the heating of solid magnesium in the presence of nitrogen gas. In a particular experiment, a 9.27-g sample of N2 completely reacts. What is the mass of magnesium must have been consumed during the reaction if the reaction yield is 100%?

Answer 1

The mass of magnesium that has been consumed, was 6.69 g

Step-by-step explanation:

The reaction is this one:

3Mg (s) + N₂(g) → Mg₃N₂

3 moles of solid magnesium react with 1 mol of nitrogen, to make 1 mol of magnesium nitride.

If 9.27 grams of nitrogen react, we see that ratio is 1:1, so we make 9.27 grams of nitride.

Mass / Molar mass = Moles

9.27 g / 100.9 g/m = 0.092 moles

If we have 0.092 moles of nitride, ratio between Mg is 1:3 so, the rule of three will be:

1 mol of Nitride was produced by 3 moles of Mg (s)

0.092 moles of nitride were produced by, (0.092 .3)/1 = 0.275 moles

Mass og Mg = 24.3 g/m

Molar mass . Moles = Mass

0.275 m . 24.3g/m = 6.69 g