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

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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] · Answer 1 · 2021-12-07T14:51:28+03:00

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.

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