Balancing chemical equations? Really need some help please!?

Question

if someone can help me out i'd be forever grateful

some of these have subscripts such as 02 (the 2 is a subscript) and N2 (the 2 is a subscript) and H20 and Mg(OH)2

1.) 2Mg (s) + O2 (g) ---> 2 __________ (s)
(magnesium) (oxygen)

2) 3 Mg (s) + N2 (g) ----> __________ (s)
(magnesium) (nitrogen)

3) __________ (s) + 6 H20 (I) ----> 3 Mg(OH)2 (s) + 2 NH3 (g)
(magnesium nitride) (water) (magnesium hydroxide) (ammonia)

4) Mg(OH)2 (s) ------> __________ (s) + H20 (g)
(magnesium hydroxide)


5) What could cause the experimental weight of oxygen to be (a) lower than expected? and (b) higher than expected? (no same answer for both)

6) Suppose the magnesium used in an experiment had a thin oxide coating. How would this affect moles of oxygen per mole of magnesium?

Thank you SO much.

Answer 1

1) 2Mg (s) + O2 (g) ---> 2 _MgO_ (s)
2) 3 Mg (s) + N2 (g) ----> __Mg3N2__ (s)
3) Mg3N2_ (s) + 6 H20 (I) ----> 3 Mg(OH)2 (s) + 2 NH3 (g)
4) Mg(OH)2 (s) ------> ___MgO____ (s) + H20 (g)

**sorry i dont know the answers to #5 + #6** the first four are correct. i took AP Chemistry, so its all good in da hood

Answer 2

1. 2Mg + O2 -> 2MgO2
2. 3Mg + N2 -> Mg3N2
3. Mg3N2 + 6H2O -> 3Mg(OH)2 + 2NH3
4. Mg(OH)2 -> MgO + H2O

5. If Mg is already oxidized - as in question 6 - then the experimental weight of O2 will be lower than expected. If the oxygen is not pure - for example contains some N2 - then the experimental weight of oxygen will be higher than expected.

6. As in question 5 you will need less oxygen.

Answer 3

1) 2Mg + O2 --->2 MgO

2) 3 Mg + N2 --->Mg3N2

3) Mg3N2 + 6H2O---> 3 Mg(OH)2 + 2NH3

4) Mg(OH)2---> MgO + H2O

5) improper measurement (you are measuring the already reacted magnesium), a coating of oxide on the magnesium

6) Some of the magnesium had already reacted with the magnesium, so it would take less O2