Determine the volume (L) of nitrogen monoxide gas that is created at STP when 32.2 g

of solid copper reacts with excess nitric acid.

3Cu(s) + 8HNO3(aq) — 3Cu(NO3)2 (aq) + 4H2O(1) + 2NO(g)

Answers

Answer 1

Taking into account the reaction stoichiometry and STP conditions, the volume of nitrogen monoxide gas that is created at STP when 32.2 g of solid copper reacts with excess nitric acid is 7.5677 L.

Reaction stoichiometry

The balanced reaction is:

3 Cu(s) + 8 HNO₃(aq) → 3 Cu(NO₃)₂ (aq) + 4 H₂O(l) + 2 NO(g)

By reaction stoichiometry, the following amounts of moles of each compound participate in the reaction:

Cu: 3 molesHNO₃: 8 molesCu(NO₃)₂: 3 molesH₂O: 4 moles NO: 2 moles

The molar mass of the compounds is:

Cu: 63.54 g/moleHNO₃: 63 g/moleCu(NO₃)₂: 187.54 g/moleH₂O: 18 g/moleNO: 30 g/mole

By reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

Cu: 3 moles× 63.54 g/mole= 190.62 gramsHNO₃: 8 moles× 63 g/mole= 504 gramsCu(NO₃)₂: 3 moles ×187.54 g/mole= 562.62 gramsH₂O: 4 moles ×18 g/mole= 72 gramsNO: 2 moles ×30 g/mole= 60 grams

STP conditions

The STP conditions refer to the standard temperature and pressure, which values are 0 °C and 1 atmosphere and are reference values for gases. In these conditions 1 mole of any gas occupies an approximate volume of 22.4 liters.

Moles of NO formed

The following rule of three can be applied: if by reaction stoichiometry 190.62 grams of Cu form 2 moles of NO, 32.2 grams of Cu form how many moles of NO?

moles of NO= (32.2 grams of Cu× 2 moles of NO)÷ 190.62 grams of Cu

moles of NO= 0.3378 moles

Then, 0.3378 moles of NO are formed.

Volume of NO created

Now, you can apply the following rule of three: if by definition of STP conditions 1 mole of NO occupies a volume of 22.4 liters, 0.3378 moles occupies how much volume?

volume= (0.3378 moles× 22.4 L)÷ 1 mole

volume= 7.5677 L

Finally, the volume of NO created is 7.5677 L.

Learn more about the reaction stoichiometry:

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Learn more:

https://brainly.com/question/10535983https://brainly.com/question/21089350

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If the initial concentration of C₂H₅OH was 1.25 × 10²² M, the half-life is 2.60 × 10⁻⁵ s.

The time required for all the 1.25 × 10²² M C₂H₅OH to decompose is 5.21 × 10⁻⁵ s.

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