Respuesta :
Answer is: 22,65 gramsof methanol must be burned.
Chemical reaction: CH₃OH + 3/2O₂ → CO₂ + 2H₂O. ΔH = -764 kJ.
Make proportion, if one mole of ethanol release 764 kJ of heat, than:
1 mol : 764 kJ = n(CH₃OH) : 541 kJ.
n(CH₃OH) = 0,708 mol.
m(CH₃OH) = n(CH₃OH) · M(CH₃OH).
m(CH₃OH) = 0,708 mol · 32 g/mol.
m(CH₃OH) = 22,65 g.
Chemical reaction: CH₃OH + 3/2O₂ → CO₂ + 2H₂O. ΔH = -764 kJ.
Make proportion, if one mole of ethanol release 764 kJ of heat, than:
1 mol : 764 kJ = n(CH₃OH) : 541 kJ.
n(CH₃OH) = 0,708 mol.
m(CH₃OH) = n(CH₃OH) · M(CH₃OH).
m(CH₃OH) = 0,708 mol · 32 g/mol.
m(CH₃OH) = 22,65 g.
[tex]\boxed{{\text{22}}{\text{.69 g}}}[/tex] of methanol must be burned to produce 541 kJ of heat.
Further explanation:
Combustion reaction:
It is the reaction in which the reactant reacts with molecular oxygen to form carbon dioxide and a water molecule. Molecular oxygen acts as the oxidizing agent in these reactions. A large amount of heat is released and therefore combustion reactions are exothermic in nature.
The change in enthalpy that occurs due to the formation of one mole of the compound from its constituent elements is called the standard enthalpy of formation. It is represented by [tex]\Delta {H_{{\text{rxn}}}}[/tex] and its units are kJ/mol.
The combustion of methanol occurs as follows:
[tex]{\text{C}}{{\text{H}}_3}{\text{OH}}+\frac{3}{2}{{\text{O}}_2}\to{\text{C}}{{\text{O}}_2}+2{{\text{H}}_{\text{2}}}{\text{O}}[/tex]
The value of [tex]\Delta {{\text{H}}_{{\text{reaction}}}}[/tex] is -764 kJ/mol. This indicates the heat produced when one mole of methanol is combusted. The negative sign means the heat is released during the process.
The number of moles of methanol [tex]\left( {{\text{C}}{{\text{H}}_3}{\text{OH}}} \right)[/tex] required in the given reaction is calculated as follows:
[tex]\begin{aligned}{\text{Moles of C}}{{\text{H}}_3}{\text{OH}}&=\left({541{\text{ kJ}}} \right)\left( {\frac{{{\text{1 mol}}}}{{764{\text{ kJ}}}}}\right)\\&=0.708{\text{ mol}}\\\end{aligned}[/tex]
The formula to calculate the mass of methanol [tex]\left( {{\text{C}}{{\text{H}}_3}{\text{OH}}} \right)[/tex] is as follows:
[tex]{\text{Mass of C}}{{\text{H}}_3}{\text{OH}}=\left({{\text{Moles of C}}{{\text{H}}_3}{\text{OH}}}\right)\left({{\text{Molar mass of C}}{{\text{H}}_3}{\text{OH}}}\right)[/tex] ...... (1)
The number of moles of [tex]{\text{ C}}{{\text{H}}_3}{\text{OH}}[/tex] is 0.708 mol.
The molar mass of [tex]{\text{ C}}{{\text{H}}_3}{\text{OH}}[/tex] is 32.04 g/mol.
Substitute these values in equation (1).
[tex]\begin{aligned}{\text{Mass of C}}{{\text{H}}_3}{\text{OH}}&=\left({{\text{0}}{\text{.708 mol}}} \right)\left({\frac{{{\text{32}}{\text{.04 g}}}}{{{\text{1 mol}}}}}\right)\\&=22.688{\text{ g}}\\&\approx {\text{22}}{\text{.69 g}}\\\end{aligned}[/tex]
Therefore the mass of methanol burned is 22.69 g.
Learn more:
1. Calculate [tex]\Delta {\text{H}}[/tex] for the reaction using Hess law: https://brainly.com/question/11293201
2. Calculate the hydroxide ion concentration: https://brainly.com/question/11293214
Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Thermodynamics
Keywords: methanol, CH3OH, combustion, 746 kJ, 541 kJ, mass of CH3OH, molar mass of CH3OH, 22.69 g, moles of CH3OH.
