hw help (balancing chemical equations)?

Question

can someone please help me balance these equations and EXPLAIN how you did it? i get confused a lot

a. C3H8 (g) + O2 (g) --> H2O (g) + CO2 (g)

b. CH4 (g) + O2 --> CO2 (g) + H2O + energy

c. AgNO3 + MgCl2 --> Mg(NO3)2 + AgCl


thank you VERYYY much!

Answer 1

C3H8 + 5 O2 ---> 4 H2O + 3 CO2

CH4 + 2 O2 ---> 2 H2O + CO2

2 AgNO3 + MgCl2 ---> Mg(NO3)2 + 2 AgCl

Answer 2

For a.
Since ther is 8 hydrogen on the reactants side, you want to balance it to 8 hydrogen on teh product side so you put a 4 in front of the H20
C3H8+O2--->4H2O+CO2
now the hydrogen is balanced with 8 hydrogen each but now there is 4 oxygen for H2O and you need to balance that later
now balance the carbon which you would put a 3 in front of CO2 to make 3 Carbon for both sides
C3H8+O2---->4H2O+3CO2
now both the hydrogen and carbon are balanced
now there is 10 oxygen on the products becasue 4*1 +3*2 =10
now just put a 5 in front of O2 and it is balanced
balanced equation: C3H8+5O2---->4H2O+3CO2
For b. it is the same thing except i don't know what energy is
For c. Mg(NO3)2 that means 1 Magnesium, 2 nitrogen, 6 oxygen
and if you put a 2 in front of Mg then it becomes 2 Mg, 4 N, 12 O and its the same concept for a
hope that helps

Answer 3

The equations need to balance, that is have the same # of C's, H's and O's on both sides.

a.) C3H8 + O2 ---> H2O + CO2

C3H8 + O2 ---> 4H2O + 3CO2 At this point we have accounted for the 8H's and 3C's on the left side. but we still need to balance the O's. We now have 10 O's on the right that we now need to account for on the left side. It looks like 5 O2's will do it.
C3H8 + 5O2 ---> 4H2O + 3CO2

b) CH4 + O2 ---> CO2 + H2O

CH4 +2O2 ---> CO2 + 2H2O


c) AgNo3 + MgCl2 ---> Mg(NO3)2 + AgCl

2AgNo3 + MgCl2 ---> Mg(NO3)2 + 2AgCl

Answer 4

In the first place, try writing all those equations with the numbers subscripted. Ok, now I assume you know how to do basic algebra. Pretend the arrow is an equals (=) sign. You want to have to same number of every element on each side.
For example, in 'a', you have three carbons (C) on the left and only one on the right. If you give it the compound CO2 the coefficient of '3', you now have the same number of carbons on each side. Leave the subscripted numbers alone, they do not change.
Now, you do have to deal with those numbers. You multiply the coefficients with the subscripts to find out how many there are. If you see no number in front, it is a '1'. And remember that '3' we just added, multiply that by the '2' to get '6' oxygens (O) on the right.

Just do that for all your compounds! If it helps, keep tally off to the side so you know how many you have on each side. Good luck!