Balance the following chemical equation by the half-reaction method in acidic conditions. Show your work. P4 + NO3–→ H2PO4– + NO(A) What is the oxidized element? What is the reduced element?
(B) Write balanced half reactions for the oxidation reaction:
(C) Write balanced half reactions for the reduction reaction:

A. Nitrate is the reduced element.

B. Phosphorus is the oxidized element.

C. oxidation half reaction:   P4 + 4H2O ⇒ H2PO4- + 6H+ + 5e

D. Reduction half reaction: NO3- +4H+ 3e- ⇒ NO + 2H2O

Explanation:

P4 + NO3–→ H2PO4– + NO(A)

nitrate is the reduced element as it has lower its oxidation state.

phosphorus is the oxidised element as it has increase in oxidation number.

B) OXIDATION REACTION:

Phosphorus has undergone oxidation from P0 to P5

P4 + 4H2O ⇒ H2PO4- + 6H+ + 5e-

C) REDUCTION REACTION:

Nitrate ion had undergone reduction from nitrate anion to nitrous oxide.

N+5 to N+2

NO3- +4H+ 3e- ⇒ NO + 2H2O

REDOX REACTION:

now multiplying oxidation reaction with 3 and the reduction reaction with 5

P4 + 5NO3- +2H+ + 2H2O⇒ 5NO+3H3PO4-

the answer

explanation:

the capillary tube was too close to the bottom of the beaker.

the ruler may have moved.

water got into the capillary tube.

the temperature was not allowed to equilibrate in the 2-4 minutes.

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%