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blondinia [14]
3 years ago
7

What is the limiting reactant if 4.0 g of Nh3 react with 8.0 g of oxygen?

Chemistry
2 answers:
Mazyrski [523]3 years ago
8 0
NH3-The limiting reactant is the reactant that get completely used up in a reaction
AlexFokin [52]3 years ago
7 0

Answer : The limiting reagent is O_2.

Solution : Given,

Mass of NH_3 = 4.0 g

Mass of O_2 = 8.0 g

Molar mass of NH_3 = 17 g/mole

Molar mass of O_2 = 32 g/mole

First we have to calculate the moles of NH_3 and O_2.

\text{ Moles of }NH_3=\frac{\text{ Mass of }NH_3}{\text{ Molar mass of }NH_3}=\frac{4.0g}{17g/mole}=0.24moles

\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{8.0g}{32g/mole}=0.25moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

4NH_3+5O_2\rightarrow 4NO+6H_2O

From the balanced reaction we conclude that

As, 5 mole of O_2 react with 4 mole of NH_3

So, 0.25 moles of O_2 react with \frac{0.25}{5}\times 4=0.20 moles of NH_3

From this we conclude that, NH_3 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.

Hence, the limiting reagent is O_2.

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