Balance the equation
1 . Fe+ H₂O → Fe₂O₃ +H₂
2 . NH₃+ O₂ → NO+H₂O
3 . NH₃ +O₂ → N₂ +H₂O
4 . C₂H₅ + O₂ → CO₂+H₂O
5 . FeCl₃ + NH₄OH → Fe (OH )₃ +NH₄Cl
6 . CH₄ +O₂→ CO₂+H₂O
7 . AlN +H₂O→ Al(OH)₃ +NH₃
8 . CuO +NH₃→ Cu+ H₂O+N₂
9 . Pb (NO₃)₂ → PbO+NO₂ +O₂
10 . Al+O₂ →Al₂O₃
11 . Be₂C + H₂O → Be (OH)₂ +CH₄
12 . MgC₂ +H₂O → Mg (0H)₂ + C₂H₂
Pls solve it's quite urgent....
Answers
Explanation:
Step 1: Tally the number of atoms based on subscript.
F
e
+
O
2
→
F
e
2
O
3
(unbalanced)
left side: Fe = 1; O = 2
right side: Fe = 2; O = 3
Step 2: Start balancing the simple molecules first, in this case the
O
2
F
e
+
3
O
2
→
F
e
2
O
3
left side: Fe = 1; O = 2 x
3
= 6
right side: Fe = 2; O = 3
Since there are now 6
O
atoms on the left, there must also be 6
O
atoms on the right. Notice that the right hand is one big molecule so whatever coefficient you would use to balance the
O
atoms, you would also need to apply to the bonded
F
e
atom.
F
e
+
3
O
2
→
2
F
e
2
O
3
left side: Fe = 1; O = 2 x 3 = 6
right side: Fe = 2 x
2
= 4; O = 3 x
2
= 6
Step 3: Now balance the rest of the equation.
4
F
e
+
3
O
2
→
2
F
e
2
O
3
left side: Fe = 1 x
4
= 4; O = 2 x 3 = 6
right side: Fe = 2 x 2 = 4; O = 3 x 2 = 6
The equation is now balanced.