Hey guys! Stuck on some Redox Reactions. I tried my best to put as much work as I could, but I'm super stuck on these ones. Thanks for any help:
1) Balance the equation as written, what is the coefficient for O2?
NH4+ + O2-->NO2− + H2O (in basic solution)
Maybe answer:
I tried treating it as an acid so I know I have to change the H+ ions at the end but I don't remember how to do that.
2(NH4+2H20->NO2-+8H+6e-)
3(4e-+O2+4H+->2H2O)
2NH4++4H2O+3O2+12H+->6H2O+2NO2-+16H+
(What do I do from here? Sorry!)
2) Balance the equation as written, what is the coefficient for Cu2+?
Cu + HNO3-->Cu2+ + NO
Maybe work:
All I know is
Cu->Cu2++2e-
I can't figure out HNO3--> Sorry!
3)Balance the equation as written, what is the coefficient for O2?
CH3CH2OH + O2-->CH3CO2H + H2O (in acidic solution)
Maybe work:
I'm confused by this one, I know it is in acid so I can use the symbolic equation 2H++"O"->H2O somewhere, but I'm super lost! Sorry!
4) A 1.5000 g sample of impure iron (II) ammonium sulfate was found to require 27.03 ml of 0.01876 M potassium permanganate (in excess acid) to reach the end point. What is the percentage of Fe in the sample?
Maybe answer:
Fe(NH4)2(SO4)2 and KMnO4
Fe2+->Fe3++e-
MnO4-+8H++5e- -->Mn2+4H2O
Balanced equation for this reaction:
5Fe2++MnO4-+8H+->5Fe3++Mn2++4H20
Moles MnO4-=.02703l*.01876M=5.07*10^-4
Moles Fe2+=5*5.07*10^-4=.002535414 mol
Mass of Fe2+=.0025mols*55.85g/mol=.1416g
Percent of Fe=.1416/1.5g*100=9.440% (my answer)
http://www.webqc.org/balance.php?reaction=Cu+%2B+HNO3%28dilute%29+%3D+Cu%28NO3%292+%2B+H2O+%2B+NO
Thank you. I got 2 and 3, but I still don't understand how to balance the first one. Can you offer any help with that? Also, if you could, could you check my last problem? I'm sorry, and thank you!
1) To balance the equation NH4+ + O2 --> NO2- + H2O in basic solution and find the coefficient for O2, you need to ensure that the number of atoms of each element is the same on both sides of the equation. Here's how you can do it step by step:
First, balance the elements other than hydrogen and oxygen. In this case, the only elements are nitrogen (N) and hydrogen (H), both appearing on both sides of the equation. Since the number of nitrogen atoms is already balanced, let's balance hydrogen next.
There are 4 hydrogen atoms on the left side (2 from NH4+ and 2 from H2O) and 2 hydrogen atoms on the right side (from H2O). To balance hydrogen, add 2 more H+ ions to the right side:
NH4+ + O2 --> NO2- + 2H2O
Now, let's balance oxygen. There are 2 oxygen atoms on the left side (from O2) and 4 oxygen atoms on the right side (2 from NO2- and 2 from H2O). To balance oxygen, add 2 more O2 molecules to the left side:
2NH4+ + O2 --> NO2- + 2H2O
Now, the equation is balanced. In the final balanced equation, the coefficient for O2 is 1.
2) To balance the equation Cu + HNO3 --> Cu2+ + NO and find the coefficient for Cu2+, you need to follow a similar process. Here's what you can do step by step:
First, balance the elements other than copper (Cu). In this case, the only other element is nitrogen (N). There is one nitrogen atom on both sides of the equation, so it is already balanced.
Now, let's balance hydrogen (H). There are 3 hydrogen atoms on the left side (from HNO3) and none on the right side. To balance hydrogen, add 3 H2O molecules to the right side:
Cu + HNO3 --> Cu2+ + NO + 3H2O
Finally, let's balance oxygen (O). There are 3 oxygen atoms on the left side (from HNO3) and 6 oxygen atoms on the right side (3 from NO and 3 from H2O). To balance oxygen, add 3 more NO molecules to the left side:
Cu + 4HNO3 --> Cu2+ + 3NO + 3H2O
Now, the equation is balanced. In the final balanced equation, the coefficient for Cu2+ is 1.
3) To balance the equation CH3CH2OH + O2 --> CH3CO2H + H2O in acidic solution and find the coefficient for O2, you can follow these steps:
First, balance the elements other than hydrogen and oxygen. Here, the only other element is carbon (C). There are 2 carbon atoms on both the left and right sides, so carbon is already balanced.
Now let's balance hydrogen (H). There are 6 hydrogen atoms on the left side (from CH3CH2OH) and 6 hydrogen atoms on the right side (3 from CH3CO2H and 3 from H2O). Hydrogen is already balanced.
Next, let's balance oxygen (O). There are 2 oxygen atoms on the left side (from O2) and 3 oxygen atoms on the right side (2 from CH3CO2H and 1 from H2O). To balance oxygen, add 1 more O2 molecule to the left side:
CH3CH2OH + O2 --> CH3CO2H + H2O
Now, the equation is balanced. In the final balanced equation, the coefficient for O2 is 1.
4) To find the percentage of Fe in a sample of impure iron (II) ammonium sulfate, you can use stoichiometry and the balanced equation given. Here's how you can calculate it:
First, write the balanced equation:
5Fe2+ + MnO4- + 8H+ -> 5Fe3+ + Mn2+ + 4H2O
Then, determine the moles of MnO4- used in the reaction. This is given as 27.03 mL of 0.01876 M potassium permanganate, so you can calculate:
Moles MnO4- = 27.03 mL * 0.01876 M = 0.507 mol
Since the balanced equation shows a 1:5 ratio between Fe2+ and MnO4-, you can determine the moles of Fe2+ used in the reaction:
Moles Fe2+ = 5 * 0.507 mol = 2.535 mol
Next, calculate the mass of Fe2+ in the sample using the molar mass of Fe:
Mass of Fe2+ = 2.535 mol * 55.85 g/mol = 141.6 g
Finally, calculate the percentage of Fe in the sample by dividing the mass of Fe2+ by the mass of the sample and multiplying by 100:
Percentage of Fe = (141.6 g / 1.5000 g) * 100 = 9.44%
So, the percentage of Fe in the sample is approximately 9.44%.