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Debora [2.8K]
3 years ago
6

A common neutralization reaction, that is used in titrations, involves sodium hydroxide, NaOH, reacting with nitric acid, HNO3.

What are the products of this reaction?
Chemistry
2 answers:
NeX [460]3 years ago
5 0

<u>Answer:</u> The products of the reaction are NaNO_3\text{ and }H_2O

<u>Explanation:</u>

Neutralization reaction is defined as the reaction when an acid reacts with a base to produce a salt and water molecule.

The chemical equation for the reaction of sodium hydroxide and nitric acid follows:

NaOH+HNO_3\rightarrow NaNO_3+H_2O

By Stoichiometry of the reaction:

1 mole of sodium hydroxide reacts with 1 mole of nitric acid to produce 1 mole of sodium nitrate and 1 mole of water.

Hence, the products of the reaction are NaNO_3\text{ and }H_2O

sertanlavr [38]3 years ago
4 0
The  common  neutralization  reaction  that  involve  NaOH  reacting   with  HNO3  produces

NaNO3     and  H2O

The  equation  for  reaction  is      as folows
NaOH  + HNO3  =  NaNO3  +  H2O 
that  is  1  mole  of  NaOH  reacted  with  1  mole  of  HNO3  to  form  1  mole  of  NaNO3  and  1  mole  of  H2O
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kumpel [21]

Answer : The limiting reactant is O_2

Explanation : Given,

Mass of C_3H_8 = 30.0 g

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Molar mass of O_2 = 32 g/mole

First we have to calculate the moles of C_3H_8 and O_2.

\text{ Moles of }C_3H_8=\frac{\text{ Mass of }C_3H_8}{\text{ Molar mass of }C_3H_8}=\frac{30.0g}{44g/mole}=0.682moles

\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{75.0g}{32g/mole}=2.34moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

C_3H_8+5O_2\rightarrow 3CO_2+4H_2O

From the balanced reaction we conclude that

As, 5 mole of O_2 react with 1 mole of C_3H_8

So, 2.34 moles of O_2 react with \frac{2.34}{5}\times 1=0.468 moles of C_3H_8

From this we conclude that, C_3H_8 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.

Therefore, the limiting reactant is O_2

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The correct answer is:
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The reaction can be described as following:
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Explanation:

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mars1129 [50]
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