I need help balancing equations
1) CH4+O2-->CO2-->+H2O
2) Hf+N2-->Hf3N4
3) Mg+H2SO4-->MgSO4+H2
4) C2H6+O2 -->CO2+H2O
5) Pb(NO3)2 + NaI--> PbI2+NaNO3
6) Fe+O2-->Fe3O4
It won't do any good for us to just give you the answers. You need to learn to do these yourself; I'm sure you want to do that too. What is you big hang up with these problems. Most look like redox equations.
1) ch4 + 2O2 =) co2 + 2h2o
2) don't know how
3) Mg + H2SO4 =) Mgso4 +h2
4) 2C2H6 + 7O2 =)4CO2 +6H20
5)PB(NO3)2 + 2NAI =)PBI2 +2NANO3
6) 3Fe + 2O2 =)FePr
Sure! I can help you balance these equations step-by-step. Here's how you can balance each equation:
1) CH4 + 2O2 → CO2 + 2H2O
To balance the carbon (C), add a coefficient of 1 in front of CO2.
To balance the hydrogen (H), add a coefficient of 2 in front of H2O.
To balance the oxygen (O), add a coefficient of 2 in front of O2.
2) 3Hf + 2N2 → Hf3N4
To balance the hafnium (Hf), add a coefficient of 3 in front of Hf3N4.
To balance the nitrogen (N), add a coefficient of 4 in front of N2.
3) Mg + H2SO4 → MgSO4 + H2
To balance the magnesium (Mg), add a coefficient of 1 in front of Mg.
To balance the hydrogen (H), add a coefficient of 2 in front of H2.
To balance the sulfur (S), add a coefficient of 1 in front of H2SO4.
4) 2C2H6 + 7O2 → 4CO2 + 6H2O
To balance the carbon (C), add a coefficient of 2 in front of C2H6 and 4 in front of CO2.
To balance the hydrogen (H), add a coefficient of 6 in front of H2O.
To balance the oxygen (O), add a coefficient of 7 in front of O2.
5) Pb(NO3)2 + 2NaI → PbI2 + 2NaNO3
To balance the lead (Pb), add a coefficient of 1 in front of Pb(NO3)2 and PbI2.
To balance the iodine (I), add a coefficient of 2 in front of NaI and PbI2.
To balance the sodium (Na), add a coefficient of 2 in front of NaNO3.
6) 4Fe + 3O2 → 2Fe3O4
To balance the iron (Fe), add a coefficient of 4 in front of Fe.
To balance the oxygen (O), add a coefficient of 3 in front of O2 and 2 in front of Fe3O4.
These are the balanced equations for the given reactions. Let me know if you need further assistance!
Balancing equations is an important skill in chemistry as it ensures that the law of conservation of mass is upheld. To balance the given equations, we need to make sure that the number of atoms of each element on both sides of the equation is the same.
1) CH4 + O2 --> CO2 + H2O
To balance this equation, we count the number of atoms for each element on both sides:
On the left side: C: 1, H: 4, O: 2
On the right side: C: 1, H: 2, O: 3
To balance the carbon (C) atoms, we place a coefficient of 2 in front of CO2:
CH4 + O2 --> 2CO2 + H2O
Next, we balance the hydrogen (H) atoms. Since there are already 4 hydrogen atoms on the left, we need to place a coefficient of 2 in front of H2O:
CH4 + O2 --> 2CO2 + 2H2O
Finally, we balance the oxygen (O) atoms. There are already 4 oxygen atoms on the right, but only 2 on the left. To balance this, we need to add a coefficient of 5/2 (or 2.5) in front of O2:
CH4 + 2.5O2 --> 2CO2 + 2H2O
2) Hf + N2 --> Hf3N4
Counting the number of atoms for each element:
On the left side: Hf: 1, N: 2
On the right side: Hf: 3, N: 4
To balance the hafnium (Hf) atoms, we place a coefficient of 3 in front of Hf:
3Hf + N2 --> Hf3N4
Now, the hafnium is balanced, but there are 6 nitrogen (N) atoms on the right side and only 2 on the left. To balance this, we place a coefficient of 2 in front of N2:
3Hf + 2N2 --> Hf3N4
3) Mg + H2SO4 --> MgSO4 + H2
Counting the number of atoms for each element:
On the left side: Mg: 1, H: 2, S: 1, O: 4
On the right side: Mg: 1, S: 1, O: 4, H: 2
Mg, S, and O are already balanced. To balance the hydrogen (H) atoms, we place a coefficient of 2 in front of H2:
Mg + H2SO4 --> MgSO4 + 2H2
4) C2H6 + O2 --> CO2 + H2O
Counting the number of atoms for each element:
On the left side: C: 2, H: 6, O: 2
On the right side: C: 1, H: 2, O: 3
To balance the carbon (C) atoms, we place a coefficient of 2 in front of CO2:
C2H6 + O2 --> 2CO2 + H2O
Next, we balance the hydrogen (H) atoms. Since there are already 6 hydrogen atoms on the left, we need to place a coefficient of 3 in front of H2O:
C2H6 + O2 --> 2CO2 + 3H2O
Finally, we balance the oxygen (O) atoms. There are already 4 oxygen atoms on the right, but only 2 on the left. To balance this, we place a coefficient of 7/2 (or 3.5) in front of O2:
C2H6 + 3.5O2 --> 2CO2 + 3H2O
5) Pb(NO3)2 + 2NaI --> PbI2 + 2NaNO3
Counting the number of atoms for each element:
On the left side: Pb: 1, N: 2, O: 6, I: 2, Na: 2
On the right side: Pb: 1, I: 2, Na: 2, N: 2, O: 6
All elements are already balanced.
6) Fe + O2 --> Fe3O4
Counting the number of atoms for each element:
On the left side: Fe: 1, O: 2
On the right side: Fe: 3, O: 4
To balance the iron (Fe) atoms, we place a coefficient of 3 in front of Fe:
3Fe + O2 --> Fe3O4
Now, the iron is balanced, but there are 4 oxygen (O) atoms on the right side and only 2 on the left. To balance this, we place a coefficient of 2 in front of O2:
3Fe + 2O2 --> Fe3O4
And that's how you balance chemical equations. By ensuring that the number of atoms of each element on both sides of the equation is the same, you maintain the law of conservation of mass.