How do you balance these equations?

Question

a.H2O ----> H2 + O2

b. NaCl + H2O

c. Fe + S ----> FeS

d. CaCo3

e. 3P4 + 10 KClO3 ----> 10 KCl + 6P2O5

f. H2SO4 + Al ----> Al2(SO4)3 + 3H2

I have been trying to solve them for homework, but I just can't get them right. Can someone explain how to solve these for me?

Thanks in advance!

and also, some of these may already be balanced, and some might not be.

Answer 1

a. 2H2O ----> 2H2 + O2

b. NaCl + H2O (?...this isn't an equation)

c. Fe + S ----> FeS (already balanced)

d. CaCo3 (this isn't an equation either, nothing to balance)

e. 3P4 + 10 KClO3 ----> 10 KCl + 6P2O5 (comes balanced)

f. 3H2SO4 + 2Al ----> Al2(SO4)3 + 3H2

Basically the trick is just to make sure the numbers of each element on each side are equivalent. A lot of times these problems are just guess and check.

Answer 2

I love these things...okay, so 2H2O--> 2H2 + O2
H-4 H-4
O-2 O-2
NaCl + H2O, this equation needs another side in order to balance/finish it- is this a net ionic(AX + BY--> AY + BX) equation?; if so then try:
-NaCl + H2O --> NaO + ClH2
-Na_aq + Cl_aq + H2O_ l --> Na_aq + O_g + Cl_aq + H2_g
-cross cancel
H2O_l--> O_g + H2_g

This is not right either- you need one more side or some numbers... Fe + S ---> FeS
CaCo3, nothing to balance here
3P4 + 10KClO3 --> 10KCl + 6P2O5 =
P-12 P-12
K-10 K-10
Cl-10 Cl- 10
O-30 O- 30
>that one was already balanced, some chem. teachers do taht, like mine, to make sure you know what you're doing- a check for any slip ups, etc.
the last one;
H2SO4 + AI --> AI2(SO4)3 + 3H2
H- 2 H- 6
S- 1 S- 3
O- 4 O- 12
AI- 1 AI- 2
= 3H2SO4 + 2AI --> Al2(SO4)3 + 3H2
H- 6 H-6
S- 3 S- 3
O- 12 O- 12
AI- 2 AI- 2
Hope this helps, good luck

Answer 3

for the first equation :
on the right hand side u have 2 hydrogen atoms and 2 oxygen
but on theleft hand side u have 2 H and 1 O
so we have to choose one of them H or O to balance first lets start with the O
the right hand side must contain 2 O so we multiply by 2
so the equation will be 2 (H2O)===H2+O2

For the other equations
NACL+H2O=====NAOH+HCL

P4+KCLO3====KCL+P2O5
for P at left =4
at right =2
so multiply by 2
P4+KClO3====KCL+2P2O5
then K at left =1
at right =1

then Cl at left=1
at right =1

then O at left=3
at right =2*5 =10
so we must multiply the left hand by 10(vlue from right)
and the right hand side by 3 (value from left)
P4+10KClO3====KCL+3(2P2O5)
P4+10KClO3====KCL+6P2O5
check again for K
10 1
SO THE FINAL EQUATION WILL BE
P4+10KClO3====10KCL+6P2O5

I Know im not very good in explaning but i tried to help

Answer 4

Where are the equations for b and d? I do not understand why you have included already balanced equations.

a. 2H2O ---> 2H2 + O2

c and e are already balanced

e.3H2SO4 + 2Al ----> Al2(SO4)3 + 3H2

Answer 5

you have to balence all the molecules like for a: you have 2 hydrgoens and 1 oygen on the left but 2hydrogens and 2 oxygens on the right, so you need to put 2H2O--->2H2 + O2
so you now have 4hygrogens and2 oxygens on the left and the same on the right
hope that helps

Answer 6

a. H2O --> H2 + 0.5O
c.i think is o.k
d.also
e.is o.k
f.3H2SO4 + 2Al --> Al2(SO4)3 + 3H2