1. Determine the amount of heat (enthalpy) produced if 25.00 grams of hexadecane is burned in...

1. Determine the amount of heat (enthalpy) produced if 25.00 grams of hexadecane is burned in excess oxygen to produce carbon dioxide and water. Ensure that your answer includes the molecular formula and a balanced chemical equation for the combustion reaction.

2. Calculate the mass of CO2 produced from 25.00 grams of hexadecane.

Expert Solution

1. Balanced chemical reaction

2C₁₆H₃₄ + 49O₂ 34H₂O + 32CO₂

one mole of hexadecane produce 10699.1 KJ heat

molar mass of hexadacane = 226.448 gm / mole that mean 1 mole of hexadacane = 226.448 gm

then 25 gm hexadecane = 125/226.448 = 0.1104 mole of hexadecane

one mole of hexadecane produce 10699.1 KJ heat then 0.1104 mole of hexadecane produce

0.110410699.1/1 = 1181.18064 KJ of heat

25 gm hexadecane produce 1181.18064 KJ of heat

2. According to reaction 2 mole of hexadecane produce 32 mole of CO₂ then 0.1104 mole of hexadecane produce

0.110432/2 = 1.7664 mole of CO₂

molar mass of CO₂ = 44.01gm/mol that mean 1 mole of CO₂ = 44.01gm then 1.7664 mole= 1.7664 44.01 =77.739 gm

25.00 grams of hexadecane produce 77.739 gm of CO₂

queen_honey_blossom answered 1 year ago

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