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Tema [17]
3 years ago
7

Suppose that 13 mol NO2 and 3 mol H2O combine and react completely. How many moles of the reactant in excess are present after t

he reaction has completed
Chemistry
1 answer:
strojnjashka [21]3 years ago
7 0

Answer:

The number of moles of excess reagent NO₂ that are present after the reaction has completed is 7 moles.

Explanation:

The balanced reaction is:

3 NO₂ + H₂O → 2 HNO₃ + NO

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reactants and products participate in the reaction:

  • NO₂: 3 moles
  • H₂O: 1 mole
  • HNO₃: 2 moles
  • NO: 1 moles

The limiting reagent is one that is consumed in its entirety first, determining the amount of product in the reaction. When the limiting reagent ends, the chemical reaction will stop.

In other words, the limiting reagent is that reagent that is consumed first in a chemical reaction, determining the amount of products obtained. The reaction depends on the limiting reagent, because the other reagents will not react when one is consumed.

You can apply the following rule of three: if by stoichiometry of the reaction 3 moles of NO₂ react with 1 mole of H₂O, 13 moles of NO₂ react with how many moles of H₂O?

moles of H_{2}O=\frac{13 moles of NO_{2}*1 mole of H_{2}O }{3 moles of NO_{2}}

moles of H₂O= 4.33 moles

But 4.33 moles of H₂O are not available, 3 moles are available. Since you have less moles than you need to react with 13 moles of NO₂, water H₂O will be the limiting reagent.

To determine the number of moles of excess reagent NO2 that are present after the reaction is complete, you can apply the following rule of three: if by stoichiometry of the reaction 1 moles of H₂O react with 3 mole of NO₂, 3 moles of H₂O react with how many moles of NO₂?

moles of NO_{2}=\frac{3 moles of NO_{2}*3 mole of H_{2}O }{1 mole of H_{2}O}

moles of NO₂= 6 moles

If 6 moles of NO₂ react and 13 moles of the compound are present, the amount that remains in excess is calculated as: 13 moles - 6 moles= 7 moles

<u><em>The number of moles of excess reagent NO₂ that are present after the reaction has completed is 7 moles.</em></u>

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