Question

A 4.5 g piece of magnesium ribbon undergoes combustion in air to produce a mixture of two ionic solids, MgO and Mg3N2. Water is added to this mixture. It reacts with the magnesium oxide to form 9.97 g of magnesium hydroxide. Assume all reactions go to completion.
A) write a balanced equation
B) how many grams of MgO are formed by the combustion of magnesium
C) how many grams of magnesium nitride are formed?

Answer 1

2Mg + O2 ==> 2MgO
3Mg + N2 ==> Mg3N2

If all of the 4.5 g Mg were converted to MgO, how much MgO would be formed?
4.5/24.305 = 0.185 moles Mg.
moles MgO = 0.185
grams MgO = 0.185 mols x molar mass MgO = 7.45 g

There were 9.97 g Mg(OH)2 recovered. How many g MgO is that.
moles Mg(OH)2 = 9.97/58.31 =0.171 moles
moles MgO = 0.171
g MgO = moles x molar mass = 0.171 x 40.305 = 6.89 g MgO.
So the difference between 7.46 and 6.89 must be the mass Mg3N2 formed.

Answer 2

There were 9.97 g Mg(OH)2 recovered. How many g MgO is that.
moles Mg(OH)2 = 9.97/58.31 =0.171 moles
moles MgO = 0.171
g MgO = moles x molar mass = 0.171 x 40.305 = 6.89 g MgO.
So the difference between 7.46 and 6.89 must be the mass Mg3N2 formed.

<b>The difference of 7.46 and 6.89 gives the difference in MgO due to Mg3N2 (not the mass of Mg3N2). So the 0.57 g difference needs to be converted to g Mg3N2.
0.57/40.305 = mols MgO = 0.0141
moles Mg3N2 = 3 x moles MgO = 3 x 0.0141 = 0.0424 and that time molar mass Mg3N2 will give grams Mg3N2.</b>

Answer 3

10.4