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What is the change in energy, in kJ, when 45.3 grams of methanol, CH3OH, combusts? 2\text{CH}_3\text{OH}\left(l\right) + 3\text{O}_2\left(g\right)\rightarrow2\text{CO}_2\left(g\right)+4\text{H}_2\text{O}\left(g\right)\hskip .5in \Delta\text{H}=-726\text{ kJ}2 CH 3 OH ( l ) + 3 O 2 ( g ) → 2 CO 2 ( g ) + 4 H 2 O ( g ) Δ H = − 726 kJ Group of answer choices -513 kJ +2,050 kJ -1,030 kJ -2,050 kJ +513 kJ

Respuesta :

Answer: -1,030 kJ

Explanation:

To calculate the number of moles we use the equation:

[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar mass}}[/tex]     .....(1)

Putting values in equation 1, we get:

[tex]\text{Moles of methanol}=\frac{45.3g}{32g/mol}=1.42mol[/tex]

The balanced chemical reaction is:

[tex]CH_3OH(l)+3O_2(g)\rightarrow 2CO_2(g)+4H_2O(g)[/tex]  [tex]\Delta H=-726kJ[/tex]

Given :

Energy released when 1 mole of [tex]CH_3OH[/tex] combusts = 726k J

Thus Energy released when 1.42 moles of [tex]CH_3OH[/tex] combusts =  [tex]\frac{726kJ}{1}\times 1.42=1030J[/tex]

Thus 1030 kJ of heat is released and as [tex]\Delta H[/tex] is negative for exothermic reaction, [tex]\Delta H=-1030kJ[/tex]

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