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3 years ago

The formula for the nitrate ion: NO3 NO2 NO3+

Chemistry

2 answers:

Leviafan [203]3 years ago

5 0

no3-Edg 2020

Explanation

Brainliest

tekilochka [14]3 years ago

3 0

Answer:

NO3-

Explanation:

edge2020

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Fossil fuels have increased the amount of crops harvested.

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3 years ago

Calculate the amount of heat released in the combustion of 10.5 grams of Al with 3 grams of O2 to form Al2O3(s) at 25°C and 1 at

ElenaW [278]

Answer : The amount of heat released in the combustion is, 209.5 kJ

Explanation :

First we have to calculate the moles of Al and $O_2$ .

$\text{ Moles of }Al=\frac{\text{ Mass of }Al}{\text{ Molar mass of }Al}=\frac{10.5g}{27g/mole}=0.389moles$

$\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{3g}{32g/mole}=0.188moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction will be:

$4Al+3O_2\rightarrow 2Al_2O_3$

From the balanced reaction we conclude that

As, 3 mole of $O_2$ react with 4 mole of $Al$

So, 0.188 moles of $O_2$ react with $\frac{4}{3}\times 0.188=0.251$ moles of $Al$

From this we conclude that, $Al$ 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 $Al_2O_3$

From the reaction, we conclude that

As, 3 mole of $O_2$ react to give 2 mole of $Al_2O_3$

So, 0.188 moles of $O_2$ react to give $\frac{2}{3}\times 0.188=0.125$ moles of $Al_2O_3$

Now we have to calculate the amount of heat released in the combustion.

As, 1 mole of $Al_2O_3$ releases amount of heat = 1676 kJ

So, 0.125 mole of $Al_2O_3$ releases amount of heat = $0.125\times 1676kJ=209.5kJ$

Thus, the amount of heat released in the combustion is, 209.5 kJ

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Read 2 more answers

The formula for the nitrate ion: NO3 NO2 NO3+​

The formula for the nitrate ion: NO3 NO2 NO3+