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%?

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

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

the correct answer is option d

explanation:

i was wrong its d because they dont use cm when calculating square roots

the positions which points up in upward carbon and down in downward carbon are axial

and vice versa holds true for equatorial positions

so the correct answer with labelled diagram will be