Chemistry: Amount of Substance and Excess Oxygen

Amount of Substance
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Amount of Substance
Question (b)
Phosphoric (V) acid + Sodium Hydroxide → Trisodium Phosphate + Water
H3PO4 (aq) + 3NaOH (aq) → Na3PO4 (aq) + 3H20 (l)
Reaction: 3H+ (aq) + (PO4)3- (aq) + 3Na+ (aq) + 3OH- (aq) → 3Na+ (aq) + (PO4)3- (aq) + 3H20 (l)
Ionic Equation: H+ (aq) + OH- (aq) → H20 (l)
Moles of sodium hydroxide contained in dissolve sodium hydroxide solution;
Moles=Mass of NaOH Molar Mass of NaOH
Moles=1g23+16+1g
= 0.025 moles
But, the mole ratio between Phosphoric (V) acid and Sodium Hydroxide is 1: 3
Therefore, moles of H3PO4 (aq) required for complete reaction = 0.025 ÷ 3
= 0.008333 moles
Question (c)
The term "excess oxygen" means that oxygen gas was supplied in abundance to allow for the complete combustion of iron (ii) sulphide. Thus, there is none of the iron (ii) sulphide that remained unburned in the reaction. This explanation is also true according to Mayer et al. (2018), who argue that the concept of excess oxygen is vital in the understanding of the theoretical or ideal yield of products in a typical chemical equation involving combustion. For example, in this case, the amount of oxygen allowed was sufficient for complete combustion of the 310g of the iron (ii) sulphide, and that not particles of it remained unburned.
4FeS (s) + 7O2 (g) → 2Fe2O3 (s) + 4SO2 (g)
Mole ratio 4:7:2:4
Moles of iron (ii) sulphide is given by;
Moles=Mass of FeSMolar Mas of FeS
Moles=310 g(55.8+32.1)
= 3.527 moles
Therefore, moles of iron (iii) oxide =3.527 ×24
= 1.763 moles
But, mass of iron (iii) oxide is given by; moles × molar mass
And molar mass for iron (iii) oxide = (55.8 × 2) + (16 × 3)
= 159.6
Mass of iron (iii) oxide = 1.763 × 159.6
= 281.3748
≈ 281.4g, which is the theoretical yield
Percentage yield =Actual YieldTheoritical Yield ×100
Percentage yield of iron (iii) oxide=255g281.4g ×100
= 90.618%
≈ 90.62%
Question (d)
Chemical equation for decomposition of sodium azide, NaN3;
2NaN3 (s) →2 Na (s) +3 N2 (g)
4.23kg of sodium azide equals 4,230 g
Moles of sodium azide=4,230g23+(3 ×14)
Moles here equals 6...