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?

Answer : The mass of nitrogen dioxide is, 17.6 grams

The formula of limiting reagent is,

The mass of excess reagent remains is, 3.74 grams

Explanation : Given,

Mass of = 11.5 g

Mass of = 9.91 g

Molar mass of = 30 g/mol

Molar mass of = 32 g/mol

First we have to calculate the moles of and .

and,

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

From the balanced reaction we conclude that

As, 2 mole of react with 1 mole of

So, 0.383 moles of react with moles of

From this we conclude that, is an excess reagent because the given moles are greater than the required moles and 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.

Molar mass of = 32 g/mole

Now we have to calculate the moles of

From the reaction, we conclude that

As, 2 mole of react to give 2 mole of

So, 0.383 mole of react to give 0.383 mole of

Now we have to calculate the mass of

Molar mass of = 46 g/mole

the role of the microscope world in creating new materials .

the answer is c

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