# Chemistry

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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?

• Chemistry -

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.

• error correction Chemistry -

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.

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.

• Chemistry -

10.4

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