Question

If 3.289 x 1023 atoms of potassium react with excess water, how many liters of hydrogen gas would be produced at STP? (HINT: Watch Significant Figures and Rounding.)
2 K + 2 H2O à 2 KOH + H2

Answer 1

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,

$n_{H_{2} } = 3.289 x 10^{23} atoms_{K} x \frac{1 mol_{K} }{6.0221x 10^{23} atoms_{K}} x \frac{1 mol_{H_{2}}}{2 mol_{K}}$

→ $n_{H_{2} } = 0.2731 mol_{H_{2} }$

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,

$V_{H_{2} } = \frac{n_{H_{2} } R T }{P} = \frac{0.2731 mol x 0.082057 \frac{atm L}{mol K} x 273.15 K}{1 atm}$

→  $V_{H_{2} } = 6.20 L$

So, the liters of hydrogen gas that are produced at STP by 3.289ₓ10²³ atoms of potassium are 6.20 L