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Serhud [2]
3 years ago
9

For this reaction, 11.5 g nitrogen monoxide reacts with 9.91 g oxygen gas. nitrogen monoxide (g) + oxygen (g) nitrogen dioxide (

g) What is the maximum mass of nitrogen dioxide that can be formed? g What is the FORMULA for the limiting reagent? What mass of the excess reagent remains after the reaction is complete?
Chemistry
1 answer:
Artist 52 [7]3 years ago
8 0

Answer : The mass of nitrogen dioxide is, 17.6 grams

The formula of limiting reagent is, NO

The mass of excess reagent remains is, 3.74 grams

Explanation : Given,

Mass of NO = 11.5 g

Mass of O_2 = 9.91 g

Molar mass of NO = 30 g/mol

Molar mass of O_2 = 32 g/mol

First we have to calculate the moles of NO and O_2.

\text{Moles of }NO=\frac{\text{Given mass }NO}{\text{Molar mass }NO}

\text{Moles of }NO=\frac{11.5g}{30g/mol}=0.383mol

and,

\text{Moles of }O_2=\frac{\text{Given mass }O_2}{\text{Molar mass }O_2}

\text{Moles of }O_2=\frac{9.91g}{32g/mol}=0.309mol

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

2NO(g)+O_2(g)\rightarrow 2NO_2(g)

From the balanced reaction we conclude that

As, 2 mole of NO react with 1 mole of O_2

So, 0.383 moles of NO react with \frac{0.383}{2}=0.192 moles of O_2

From this we conclude that, O_2 is an excess reagent because the given moles are greater than the required moles and NO is a limiting reagent and it limits the formation of product.

Moles of excess reagent remains = 0.309 - 0.192 = 0.117 mol

Now we have to calculate the mass of excess reagent remains.

\text{ Mass of excess reagent}=\text{ Moles of excess reagent}\times \text{ Molar mass of excess reagent}

Molar mass of O_2 = 32 g/mole

\text{ Mass of excess reagent}=(0.117moles)\times (32g/mole)=3.74g

Now we have to calculate the moles of NO_2

From the reaction, we conclude that

As, 2 mole of NO react to give 2 mole of NO_2

So, 0.383 mole of NO react to give 0.383 mole of NO_2

Now we have to calculate the mass of NO_2

\text{ Mass of }NO_2=\text{ Moles of }NO_2\times \text{ Molar mass of }NO_2

Molar mass of NO_2 = 46 g/mole

\text{ Mass of }NO_2=(0.383moles)\times (46g/mole)=17.6g

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