I really need help for my homework on balancing equations please show all the work including the coeffients if needed.
I need help with 3 questions,
1) NaN3 -> Na + N2
2) NO2 + O2 -> NO + 03
3) LiOH + CO2 -> Li2CO3 + H2O
1)
Well, to make those nitrogen atoms come out, we will need the least common multiple of 2 and 3 which is 6, so try 5 nitrogen atoms on each side. That means 2 of NaN3 and 3 of N2, then work on the sodium
2 NaN3 --> ? Na + 3 N2
I guess we can all see that our question mark is really 2
2)
I have 1 Nitrogen on the left and one on the right
I have 4 Oxygens on the left and 4 on the right
That will do as is
3)
I have 1 Li left and 2 on the right so try 2 LiOh
2LiOH + CO2 -> Li2CO3 + H2O
I now have 2 H atoms left and 2 right -- ok
I now have 4 O atoms left and 4 right-- ok
I now have 1 C atom left and one right -- ok
so I am finished
1)
Well, to make those nitrogen atoms come out, we will need the least common multiple of 2 and 3 which is 6, so try 66 nitrogen atoms on each side.
Or maybe just 6
Sure, I'd be happy to help you with balancing these chemical equations. Balancing equations involves making sure that there are an equal number of atoms of each element on both sides of the equation.
Let's start with the first equation:
1) NaN3 -> Na + N2
To balance this equation, we need to make sure that there is an equal number of each atom on both sides.
On the left side, we have 1 Na atom, 1 N atom, and 3 atoms of O. On the right side, we have 1 Na atom and 2 N atoms.
To balance the Na atoms, we need a coefficient of 2 in front of Na on the right side:
NaN3 -> 2Na + N2
Now, we have 1 N atom on the left side and 2 N atoms on the right side. To balance the N atoms, we need a coefficient of 2 in front of N2 on the right side:
NaN3 -> 2Na + 2N2
The equation is now balanced.
Moving on to the second equation:
2) NO2 + O2 -> NO + 03
On the left side, we have 2 O atoms and 2 N atoms. On the right side, we have 2 O atoms and 1 N atom.
To balance the N atoms, we need a coefficient of 2 in front of NO on the right side:
NO2 + O2 -> 2NO + 03
Now, we have 2 O atoms on the left side and 3 O atoms on the right side. To balance the O atoms, we need a coefficient of 3/2 (or 1.5) in front of O3 on the right side:
NO2 + O2 -> 2NO + 3/2O3
However, coefficients in chemical equations must be whole numbers, so we need to multiply all coefficients by 2 to get rid of the fractions:
2NO2 + 2O2 -> 4NO + 3O3
The equation is now balanced.
Lastly, the third equation:
3) LiOH + CO2 -> Li2CO3 + H2O
On the left side, we have 1 Li atom, 1 C atom, 1 O atom, and 1 H atom. On the right side, we have 2 Li atoms, 1 C atom, 3 O atoms, and 1 H atom.
To balance the Li atoms, we need a coefficient of 2 in front of LiOH on the left side:
2LiOH + CO2 -> Li2CO3 + H2O
Now, we have 2 O atoms on the left side and 3 O atoms on the right side. To balance the O atoms, we need a coefficient of 3/2 (or 1.5) in front of H2O on the right side:
2LiOH + CO2 -> Li2CO3 + 3/2H2O
Again, we need to multiply all coefficients by 2 to get rid of the fractions:
4LiOH + 2CO2 -> 2Li2CO3 + 3H2O
The equation is now balanced.
I hope this explanation helps you understand how to balance chemical equations. Let me know if you have any other questions!