Answer:
N2H2(aq) + 2OH^-(aq) ----------> N2(g) + 2H2O(l) + 2e
Explanation:
Hydrazine is mostly used in thermal engineering as an anticorrosive agent. Hydrazine can be oxidized in aqueous solution as shown in the equation above. Oxidation has to do with loss of electrons and increase in oxidation number.
The oxidation number of nitrogen in the equation increased from -1 in hydrazine on the lefthand side of the reaction equation to zero in nitrogen on the right hand side of the reaction equation. Two electrons were lost in the process as shown.
start the balancing by writing down how many atoms there are per element. we’ll use this as an example:
C3H8 + O2 --> H2O + CO2
C = 3 C = 1
H = 8 H = 2
O = 2 O = 3
balance the carbon first, as it is easiest to do. add a coefficient to the single carbon atom on the right of the equation to balance it with the 3 carbon atoms on the left of the equation:
C3H8 + O2 --> H2O + (3)CO2
now there are 3 carbon atoms on each side. however, when you do this, you multiply the amount of oxygen atoms you had. therefore, now, there are 6 carbon atoms in 3CO2, plus that other oxygen atom in H2O. you now have 7 O atoms instead of 3.
C = 3 C = 3
H = 8 H = 2
O = 2 O = 7
now let’s move on to the hydrogen atoms.
C3H8 + O2 --> H2O + 3CO2
you have 8 hydrogen atoms on the left side, and 2 on the right. in order to balance them, you have to multiply the right side’s hydrogen atoms by 4. 4(2) = 8.
C3H8 + O2 --> (4)H2O + 3CO2
now both hydrogen and carbon atoms are balanced. same amount on both sides. however, your oxygen atoms have changed due to the multiplying (right side). you now have 10 of them.
C = 3 C = 3
H = 8 H = 8
O = 2 O = 10
now we balance the oxygen atoms. multiply the left side of the equation’s oxygen atoms by 5. 5(2) = 10
C3H8 + (5)O2 --> 4H2O + 3CO2
the chemical equation is all balanced. basically, just multiply with numbers until it equals the same amount on both sides.
C = 3 C = 3
H = 8 H = 8
O = 10 O = 10
We calculate first for the number of moles of gases in the sample through the ideal gas equation.
n = PV/RT
n = (725 mmHg/760 mmHg/atm)(0.255 L) / (0.0821 L.atm/mol.K)(65 + 273.15)
n = 8.76 x 10^-3 mol
Then, we calculate for the mol N2 using the ratio of the pressure.
n N2 = (8.76 x 10^-3 mols)(231 mmHg/725 mmHg)
n N2 = 2.79 x 10^-3 moles
Then, multiply the value with the molar mass of N2 which is 28 grams per mol giving us the answer of 0.078 grams.
