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

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,

→

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,

→  

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

i think   the answer is c hope i'm right