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?
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.
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.
10.4
A) To write a balanced equation, we first need to know the oxidation states of the elements involved. In this case, magnesium (Mg) has an oxidation state of +2, oxygen (O) has an oxidation state of -2, and nitrogen (N) has an oxidation state of -3.
The combustion of magnesium produces magnesium oxide (MgO) and magnesium nitride (Mg3N2). The balanced equation for the combustion of magnesium is:
2 Mg + O2 → 2 MgO
The reaction between magnesium oxide and water produces magnesium hydroxide (Mg(OH)2). The balanced equation for this reaction is:
MgO + H2O → Mg(OH)2
B) To find the number of grams of MgO formed by the combustion of magnesium, we need to use stoichiometry. In the balanced equation, we see that 2 moles of magnesium (Mg) react to form 2 moles of MgO. So, the molar ratio between Mg and MgO is 2:2.
Given that the mass of magnesium ribbon is 4.5 g, we can calculate the number of moles of Mg using its molar mass (24.31 g/mol):
moles of Mg = mass of Mg / molar mass of Mg
= 4.5 g / 24.31 g/mol
= 0.185 moles
Using the molar ratio, we can determine the moles of MgO formed:
moles of MgO = moles of Mg
= 0.185 moles
To convert moles to grams, we multiply the moles of MgO by its molar mass (40.31 g/mol):
mass of MgO = moles of MgO × molar mass of MgO
= 0.185 moles × 40.31 g/mol
= 7.46 g
Therefore, 7.46 grams of MgO are formed by the combustion of magnesium.
C) To find the grams of magnesium nitride (Mg3N2) formed, we need to use stoichiometry. In the balanced equation, we see that 2 moles of magnesium (Mg) react to form 1 mole of Mg3N2. So, the molar ratio between Mg and Mg3N2 is 2:1.
We already know the moles of Mg from previous calculations (0.185 moles). Using the molar ratio, we can determine the moles of Mg3N2 formed:
moles of Mg3N2 = moles of Mg / 2
= 0.185 moles / 2
= 0.093 moles
To convert moles to grams, we multiply the moles of Mg3N2 by its molar mass (100.95 g/mol):
mass of Mg3N2 = moles of Mg3N2 × molar mass of Mg3N2
= 0.093 moles × 100.95 g/mol
= 9.26 g
Therefore, 9.26 grams of magnesium nitride are formed.