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Select the correctly balanced chemical equation for the reactants given:Decane (C10H22) reacts with oxygen gasa.2 C10H22 + 62 O(g) --> 20 CO2(g) + 22 H2O(g)b.C10H22 + O2(g) --> CO2(g) + H2O(g)c.C10H22 + 31 O(g) --> 10 CO2(g) + 11 H2O(g)d.C10H22 + 31 O2(g) --> 10 CO2(g) + 11 H2O(g)e.2 C10H22 + 31 O2(g) --> 20 CO2(g) + 22 H2O(g)

Question

Select the correctly balanced chemical equation for the reactants given:Decane (C10H22) reacts with oxygen gasa.2 C10H22 + 62 O(g) --> 20 CO2(g) + 22 H2O(g)b.C10H22 + O2(g) --> CO2(g) + H2O(g)c.C10H22 + 31 O(g) --> 10 CO2(g) + 11 H2O(g)d.C10H22 + 31 O2(g) --> 10 CO2(g) + 11 H2O(g)e.2 C10H22 + 31 O2(g) --> 20 CO2(g) + 22 H2O(g)

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Solution

The correctly balanced chemical equation for the reactants given is:

2 C10H22 + 31 O2(g) --> 20 CO2(g) + 22 H2O(g)

So, the correct answer is e.

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Write down a balanced equation for formation reaction of butane, 𝐶4𝐻10 (2 marks)(c) Given the following reactions and enthalpiesC4H8(l) + 6O2(g) → 4CO2(g) + 4H2O(l) ∆𝑐𝑜𝑚𝑏.𝐻298𝐾𝜃 = -2,718.8 kJ mol-1C4H8(l) + H2(g) → C4H10(l) ∆ℎ𝑦𝑑.𝐻298𝐾𝜃 = -126.8 kJ mol-1C(s, graphite) + O2(g) → CO2(g) ∆𝑓𝐻298𝐾𝜃 (𝐶𝑂2(𝑔)) = -393.5 kJ mol-1H2(g) + 12 O2(g) → H2O(l) ∆𝑓𝐻298𝐾𝜃 (𝐻2𝑂(𝑙)) = -285.8 kJ mol-1calculate the enthalpy change of formation ∆𝑓𝐻298𝐾𝜃 𝐶4𝐻10(𝑙)

. Determine the mass of carbon dioxide produced when 0.85 grams of butane (C4 H10 ) reacts with oxygen according to the following balanced chemical equation: 2 C4 H10 + 13 O2 → 8 CO2 + 10 H2 O

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