If you have 2.1x10^23 atoms of Aluminum, then what pressure (in atm) of hydrogen gas at 10C in a 7.50 L container can be made?

Question

Balanced Equation: 6 HNO₃ + 2 AL ---> 2 AL(NO₃)₃ + 3 H₂

Answer 1

6HNO₃ + 2Al ---> 2Al(NO₃)₃ + 3H₂ ... balanced equation (note that aluminum is Al and not AL)

moles of Al present = 2.1x10²³ atoms x 1 mol/6.02x10²³ atoms = 0.349 moles Al

moles of H2 that can be produced = 0.349 mol Al x 3 mol H2 / 2 mol Al = 0.524 moles H2

To convert this number of moles of H2 to pressure under the given conditions, we use the ideal as law...

PV = nRT and solving for P, we have...

P = nRT/V

P = (0.524 mol)(0.0821 Latm/Kmol)(283K)/7.50 L (note to change º C to K)

P = 1.62 atm = 1.6 atm (to 2 significant figures)