(i) 200.0 g of air was heated by the energy from the complete combustion of 1.00 mol phosphine. Calculate the temperature rise using section 1 of the data  booklet and the data below.

Standard enthalpy of combustion of phosphine,
Specific heat capacity of air = 1.00Jg⁻¹K⁻¹ = 1.00 kJkg⁻¹K⁻¹

(ii) The oxide formed in the reaction with air contains 43.6 % phosphorus by mass. Determine the empirical formula of the oxide, showing your method.

(iii) The molar mass of the oxide is approximately 285gmol⁻¹. Determine the molecular formula of the oxide.

(i) State the equation for the reaction of this oxide of phosphorus with water.

(ii) Predict how dissolving an oxide of phosphorus would affect the pH and electrical conductivity of water.

pH:

Electrical conductivity:

(iii) Suggest why oxides of phosphorus are not major contributors to acid deposition.

(iv) The levels of sulfur dioxide, a major contributor to acid deposition, can be minimized by either pre-combustion and post-combustion methods. Outline one technique of each method.

Pre-combustion:

Post-combustion:

(i)
temperature rise «\(\frac{{750 \times 1.00}}{{0.2000 \times 1.00}}\)= 3750 «°C/K»

Do not accept −3750.

(ii)
n(P)«=\(\frac{{43.6}}{{30.97}}\)»=1.41«mol»
n(O)«=\(\frac{{100 - 43.6}}{{16.00}}\)»= 3.53«mol»
\(\frac{{n\left( {\rm{O}} \right)}}{{n\left( {\rm{P}} \right)}} = \frac{{3.53}}{{1.41}} = 2.50\) so empirical formula is» P₂O₅

Accept other methods where the working is shown.

(iii)
\(\frac{{285}}{{141.9}}\)=2.00, so molecular formula=2×P₂O₅=»P₄O₁₀

(i)
P₄O₁₀ (s) + 6H₂O (l) → 4H₃PO₄ (aq)