Respuesta :
We must first write the reaction in question (be sure that the reaction is balanced),
2K + 2H₂O → 2KOH + H₂
As you can see, 2 moles of K produce 1 mol of hidrogen gas (H₂)
The Avogadro constant or Avogadro's number (NA) is the number of constituent particles (usually atoms or molecules) that are found in the amount of a mole substance, and it is equal to 6.0221ₓ10²³ particles/mol.
We are going to use the Avogadro´s number and the stequiometric coefficients of the above reaction to calculate the number of moles of hydrogen gas that are produced for 3.289ₓ10²³ atoms of potassium,
[tex]n_{H_{2} } = 3.289 x 10^{23} atoms_{K} x \frac{1 mol_{K} }{6.0221x 10^{23} atoms_{K}} x \frac{1 mol_{H_{2}}}{2 mol_{K}}[/tex]
→ [tex]n_{H_{2} } = 0.2731 mol_{H_{2} }[/tex]
According to the IUPAC, STP (Standard conditions for Temperature and Pressure) is defined as a temperature of 273.15 K and an absolute pressure of exactly 1 bar which is equal to 0.986923 atm .
To know the volume of H₂ that 0.2731 moles of this substance represents at STP we are going to use the law of ideal gases, assuming that in these conditions H₂ behave as an ideal gas.
According to the law of ideal gases:
PV = nRT → V = nRT / P
where P, V, n and T are the pressure, volume, moles and temperature of the gas in question while R is the gas constant (0.082057 atm L / mol K)
Then,
[tex]V_{H_{2} } = \frac{n_{H_{2} } R T }{P} = \frac{0.2731 mol x 0.082057 \frac{atm L}{mol K} x 273.15 K}{1 atm}[/tex]
→ [tex]V_{H_{2} } = 6.20 L[/tex]
So, the liters of hydrogen gas that are produced at STP by 3.289ₓ10²³ atoms of potassium are 6.20 L
Excess reagents are chemicals that are not completely consumed in a reaction. At STP volume of the hydrogen gas produced is 6.20 L.
What is Avogadro's number?
Avogadro's number or constant is a specific number that includes the molecules and the atoms of the compounds and is equal to [tex]6.0221 \times 10^{23}\;\rm particles/mol.[/tex]
The balanced chemical reaction can be shown as:
[tex]\rm 2K + 2H_{2}O \rightarrow 2KOH + H_{2}[/tex]
From the reaction, it can be said that 1 mole of potassium is used to produce 1 mole of hydrogen gas.
Moles of hydrogen produced is calculated using the Avogadro's number as:
[tex]\begin{aligned}\text{Moles of H}_{2} &= 3.289 \times 10^{23} \times \dfrac{1}{6.0233\times 10^{23}}\times \dfrac{1}{2}\\\\&= 0.2731\;\rm mol\end{aligned}[/tex]
At STP condition the pressure will be 1 atm and the temperature will be 273.15 K for 0.2731 mol of hydrogen gas.
The volume is calculated using the ideal gas equation as:
[tex]\begin{aligned}\rm V &= \rm \dfrac{nRT}{P}\\\\&= \dfrac{0.2731\times 0.082057\times 273.15}{1}\\\\&= 6.20\;\rm L\end{aligned}[/tex]
Therefore, at STP the volume of hydrogen gas is 6.20 L.
Learn more about volume and STP here:
https://brainly.com/question/14390346
