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Determine the amount of heat (in kJ) associated with the production of 1.76 × 10^4 g of NO2 according to the following equation:
2NO (g) + O2 (g) → 2NO2(g) ΔH = −114.6 kJ/mol
Enter your answer in scientific notation

Respuesta :

Answer: The amount of heat associated with the production of given amount of nitrogen dioxide is [tex]2.2\times 10^4kJ[/tex]

Explanation:

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

Given mass of nitrogen dioxide gas = [tex]1.76\times 10^4g=17600g[/tex]

Molar mass of nitrogen dioxide gas = 46 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of nitrogen dioxide gas}=\frac{17600g}{46g/mol}=382.60mol[/tex]

For the given chemical reaction:

[tex]2NO(g)+O_2(g)\rightarrow 2NO_2(g);\Delta H_{rxn}=-114.6kJ/mol[/tex]

By Stoichiometry of the reaction:

2 moles of nitrogen dioxide produces 114.6 kJ of heat.

So, 382.60 moles of nitrogen dioxide will produce = [tex]\frac{114.6kJ}{2}\times 382.60=21922.98kJ[/tex] of heat.

Scientific notation is defined as the notation in which a number is expressed in the decimal form. That means always written in the power of 10 form.

The amount of heat expressed in scientific notation is [tex]2.2\times 10^4kJ[/tex]

Hence, the amount of heat associated with the production of given amount of nitrogen dioxide is [tex]2.2\times 10^4kJ[/tex]

Lanuel

The amount of heat (in kJ) associated with the production of [tex]1.76 \times 10^4 \; grams[/tex] of [tex]NO_2[/tex] is equal to 21,918.96 kJ.

Given the following data:

  • Change in internal energy = -114.6 kJ/mol
  • Mass of [tex]NO_2[/tex] = [tex]1.76 \times 10^4 \; grams[/tex]

To determine the amount of heat (in kJ) associated with the production of this mass of gas:

The balanced chemical reaction is:

                        [tex]2NO_{(g)} + O_2_{(g)} --> 2NO_2_{(g)}[/tex]

First of all, we would determine the number of moles of in the above chemical reaction.

[tex]Number\;of\;moles \;(NO_2)= \frac{Mass\; of\;NO_2}{Molar\;mass\;of\;NO_2}[/tex]

Molar mass of [tex]NO_2[/tex]  = 46.01 g/mol

Substituting the parameters into the formula, we have;

[tex]Number\;of\;moles \;(NO_2)= \frac{1.76 \times 10^4}{46.01}[/tex]

Number of moles = 382.53 moles

By stoichiometry:

2 mole of [tex]NO_2[/tex] = -114.6 kJ/mol

382.53 moles of [tex]NO_2[/tex] = X kJ/mol

Cross-multiplying, we have:

[tex]2X = 382.53 \times -114.6\\\\2X = 43,837.938\\\\X = \frac{43,837.938}{2}[/tex]

X = 21,918.96 kJ

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