Balancing and writting ionic equations?

Question

My son has asked for help with his chemistry and I have no idea what to do. It was never one of my strong points at school so I was wondering if someone could help me with these, in as basic a way as you can because I really am terrible lol.

Q: Balance the following equations if necessary and write them as ionic equations.

a) HCl(aq) + NaOH(aq) --> NaCl(aq) + H2O(l)

b) HNO3(aq) + Mg(OH)2(aq) --> Mg(NO3)2(aq) + H2O(l)

c) LiCl(aq) + AgNO3(aq) --> AgCl(s) + LiNO3(aq)

d) H2SO4(aq) + LiOH(aq) --> Li2SO4(aq) + H2O(l)

e) Pb(NO3)2(aq) + KCl(aq) --> KNO3(aq) + PbCl2(s)


It's been several years since I took chemistry at school and I can't even remember what half of these symbols stand for so i'm not much help to my son unfortunately. Any help would be very mmuch appreciated. Thank You.

Answer 1

The first one is already balanced (same number of each element on both sides). I'll break down by writing as an ionic equation (notation involving writing each compound as separate ions they break into in aqueous solution-the "aq" designation):

H+ + Cl- + Na+ + OH- --> Na+ + Cl- + H2O (l)

Removing the "spectator ions," the ones that do not change from left side to right side of the equation leaves

H+ + OH- --> H2O (l)
This is also called an acid-base, or neutralization reaction, also more recently called a "metathesis reaction." In fact, all these reactions appear to be metathesis (double displacement )reactions.

Answer 2

1) get the chemical formulae of the compounds involved in the reaction - eg sodium hydroxide is NaOH 2) write out the equations in chemical symbols 3) add up how many of each symbol are on each side of the equation - a subscript number means there are that number of the symbol immediately preceeding it. you cannot change this number. 4) put normal-sized numbers in front of a compound's formula to change how many molecules of that compound are consumed/produced. you may need to change both sides of the equation to get them to balance (use the lowest common multiples you can). here is a worked example: glucose + oxygen = carbon dioxide + water 1&2) C6H12O6 + O2 = CO2 + H2O 3) left: C=6, H=12, O=(6+2)8 right: C=1, H=2, O=(2+1)3 4) if you put a 6 infront of the CO2 you will have the same number of C on each side. then you also have more O on that side, now a total of 13. we will come back to the O in a minute, but first look at the H: 12 on the left, and 2 on the right. so put a 6 infront of the H2O to make the H balance. Now C and H are balanced, and we have an even number of O on the right (18). as there are 6O on the left in glucose, we can just use 18-6=12 to balance the O. it is O2, so we need just 6O2 to make 12 O on the left. so the full balanced equation is: c6H12O6 + 6O2 = 6CO2 + 6H2O. your equations are much simpler than this example, so you should be fine. have a go! good luck...

Answer 3

Take the first one for example: HCl(aq) + NaOH(aq) --> NaCl(aq) + H2O(l)
If we addup the numbers of elements on the left and right hand sides like so:

2 H 2
1 Cl 1
1 Na 1
1 O 1

The numbers of elements must equal. In this case they do.

A different on for example could be: C3H8 + O2 --> CO2 + H20

3 C 1
8 H 2
2 O 3

Not balanced. We eventually get: C3H8 + 5O2 --> 3CO2 + 4H2O

3 C 3
8 H 8
10 O 10

where it is balanced.

Answer 4

Though the first respondent tried to balance the reactions but some mistakes were made i am only making improvements-
HCl(aq) + NaOH(aq) --> NaCl(aq) + H2O(l)

2HNO3(aq) + Mg(OH)2(aq) --> Mg(NO3)2(aq) + 2H2O(l)

LiCl(aq) + AgNO3(aq) --> AgCl(s) + LiNO3(aq)

H2SO4(aq) + 2LiOH(aq) --> Li2SO4(aq) + 2H2O(l)

Pb(NO3)2(aq) + 2KCl(aq) --> 2KNO3(aq) + PbCl2(s)

Answer 5

a) HCl(aq) + NaOH(aq) --> NaCl(aq) + H2O(l)
ok that is balanced coz there are an equal number of each element on each side

jus add up how many of each element there is on each side, and if theyre not equal then add big numbers infront of the symbol, untill you have got them all equal good luck!

Answer 6

a) HCl(aq) + NaOH(aq) --> NaCl(aq) + H2O(l)
H+ + Cl- + Na+ + OH- --> Na+ + Cl- + [H+ + OH- or H+ + O2-]
It's already balanced...because the total # of atoms of each element in both sides of the equation are the same...
__________________2-H-2
__________________1-Cl-1
__________________1-O=1
b) 2HNO3(aq) + Mg(OH)2(aq) -->Mg(NO3)2(aq) + 2H2O(l)
H+ + NO3- + Mg2+ + OH- --> Mg2+ + NO3- + H+ + O2-
_______________________4-H-4
_______________________2-N-2
_______________________8-O-8
_______________________1-Mg-1
c)LiCl(aq) + AgNO3(aq) --> AgCl(s) + LiNO3(aq)
Li+ + Cl- + Ag+ + NO3- --> Ag+ + Cl- + Li+ + NO3-
It's already balanced...because the total # of atoms of each element in both sides of the equation are the same...
___________________1-Li-1
___________________1-Cl-1
___________________1-Ag-1
___________________1-N-1
___________________3-O-3
d) H2SO4(aq) + 2LiOH(aq) --> Li2SO4(aq) + 2H2O(l)
H+ + SO4(2-) + Li+ OH- --> Li+ + SO4(2-) + H+ + O2-
_____________________4-H-4
_____________________1-S-1
_____________________6-O-6
_____________________2-Li-2
e) Pb(NO3)2(aq) + 2KCl(aq) --> 2KNO3(aq) + PbCl2(s)
Pb2+ + NO3- + K+ + Cl- --> K+ + NO3- + Pb2+ + Cl-
______________________1-Pb-1
______________________2-N-2
______________________6-O-6
______________________2-K-2
______________________2-Cl-2
To balance equation easier, balance first the nitrogen, phosphorus, and sulfur. Those elements are the most difficult to balance so if you have balance them, you can easily balance other elements...
In writing the elements in the equation to prove the balancing like this one,
1-Pb-1
2-N-2
6-O-6
2-K-2
2-Cl-2
always put the symbol of the elements below the arrow...not below the reactants nor products...

Answer 7

reactions was balanced in first answer.i give u only ionic reactions(full and short ways)


a)H+ +CL- +Na+ +OH-=Na+ +CL-+H2O
H+ +OH-=H2O

b)2H+ +2NO3-+Mg(OH)2=Mg2+ +2NO3-+2H2O
2H+ +Mg(OH)2=Mg2+ +2H20

c)LI+ +CI-+Ag+NO3-=AgCI +LI+ +NO3-
CI-+Ag+=AgCI

d)2H+ +SO4 2-+2LI+ +2OH-=2LI+ +SO4 2- +2H2O
2H++2OH-=2H2O

e)Pb2+ +2NO3- +2K+ +2CI-=2K+ +2NO3-+PbCl2
Pb2+ +2CI-=PbCI2

Answer 8

lol