Answer:
Part 1: C₉H₂₀ (l) + 14O₂ (g) ----> 9CO₂ (g) + 10H₂0 (g)
Part 2: Volume of CO₂ produced = 1223.21 L
<em>Note: the complete second part of the question is given below:</em>
<em>2. Suppose 0.470 kg of nonane are burned in air at a pressure of exactly 1 atm and a temperature of 17.0 °C. Calculate the volume of carbon dioxide gas that is produced. Round your answer to 3 significant digits.</em>
Explanation:
Part 1: Balanced chemical equation
C₉H₂₀ (l) + 14O₂ (g) ----> 9CO₂ (g) + 10H₂0 (g)
Part 2: volume of carbon dioxide produced
From the equation of the reaction;
At s.t.p., I mole of C₉H₂₀ reacts with 14 moles of O₂ to produce 9 moles of CO₂
molar mass of C₉H₂₀ = 128g/mol: molar mass of CO₂ = 44 g/mol, molar volume of gas at s.t.p. = 22.4 L
Therefore, 128 g of C₉H₂₀ produces 14 * 22.4 L of CO₂ i.e. 313.6 L of CO₂.
O.470 Kg of nonane = 470 g of nonane
470 g of C₉H₂₀ will produce 470 * (313.6/128) L of CO₂ = 1151.50 L of CO₂
Volume of CO₂ gas produced at 1 atm and 17 °C;
Using P₁V₁/T₁ = P₂V₂/T₂
V₂ = P₁V₁T₂/P₂T₁
where P₁ = 1 atm, V₁ = 1151.50 L, T₁ = 273 K, P₂ = 1 atm, T₂ = 17 + 273 = 290 K
Substituting the values; V₂ = (1 * 1151.5 * 290)/(1 * 273)
Therefore volume of CO₂ produced, V₂ = 1223.21 L of CO₂
