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3 years ago

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15. Ammonium nitrate, NH4NO3, and aluminum powder react explosively producing nitrogen gas, water vapor and aluminum oxide. Writ

e the balanced equation and calculate the enthalpy change for this reaction.

1 answer:

Lunna [17]3 years ago

8 0

Answer:

$2Al_(_S_) ~+~3NH_4NO_3_(_a_q_) ->3N_2_(_g_) ~+~6H_2O_(_g_) ~+~Al_2O_3_(_S_)$

$Entalpy=-2861.9~KJ$

Explanation:

In this case, we have to start with the <u>reagents</u>:

$Al~+~NH_4NO_3$

The compounds given by the problem are:

-) <u>Nitrogen gas</u> = $N_2$

-) <u>Water vapor</u> = $H_2O$

-) <u>Aluminum oxide</u> = $Al_2O_3$

Now, we can put the products in the <u>reaction</u>:

$Al_(_S_) ~+~NH_4NO_3_(_aq_) ->N_2_(_g_) ~+~H_2O_(_g_) ~+~Al_2O_3_(_S_)$

When we <u>balance</u> the reaction we will obtain:

$2Al_(_S_) ~+~3NH_4NO_3_(_a_q_) ->3N_2_(_g_) ~+~6H_2O_(_g_) ~+~Al_2O_3_(_S_)$

Now, for the enthalpy change, we have to find the <u>standard enthalpy values</u>:

$Al_(_S_)=0~KJ/mol$

$NH_4NO_3_(_a_q_)=-132.0~KJ/mol$

$N_2_(_g_)=0~KJ/mol$

$H_2O_(_g_)=~-~241.8~KJ/mol$

$Al_2O_3_(_S_)=~-~1675.7~KJ/mol$

With this in mind, if we <u>multiply</u> the number of moles (in the balanced reaction) by the standard enthalpy value, we can calculate the energy of the <u>reagents</u>:

$(0*2)~+~(-132*3)=~-396~KJ$

And the <u>products</u>:

$(0*3)~+~(-241.8*6)~+~(-1675.7*1)=-3125.9~KJ$

Finally, for the total enthalpy we have to <u>subtract</u> products by reagents :

$(-3125.9~KJ)-(-396~KJ)=-2729.9~KJ$

I hope it helps!

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Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CH_4+2O_2\rightarrow CO_2+2H_2O$

From the balanced reaction we conclude that

As, 2 mole of $O_2$ react with 1 mole of $CH_4$

So, 0.026 moles of $O_2$ react with $\frac{0.026}{2}=0.013$ moles of $CH_4$

From this we conclude that, $CH_4$ is an excess reagent because the given moles are greater than the required moles and $O_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $CO_2$

From the reaction, we conclude that

As, 2 mole of $O_2$ react to give 1 mole of $CO_2$

So, 0.026 moles of $O_2$ react to give $\frac{0.026}{2}=0.013$ moles of $CO_2$

Now we have to calculate the mass of $CO_2$

$\text{ Mass of }CO_2=\text{ Moles of }CO_2\times \text{ Molar mass of }CO_2$

$\text{ Mass of }CO_2=(0.013moles)\times (44g/mole)=0.572g$

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Now we have to calculate the percent yield of $CO_2$

$\% \text{ yield of }CO_2=\frac{\text{ Experimental yield of }CO_2}{\text{ Theretical yield of }CO_2}\times 100$

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From the question given above, the following data were obtained:

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Number of molecule of H₂O =?

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Next, we shall determine the mass of 1 mole of H₂O. This can be obtained as follow:

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