What volume would 60.4 grams of NH3 occupy if the temperature is 27 dgrees celcius and the pressure is 670 tor

i need the answer in liters?

Answer:
60.4 g NH3 => 60.4/17 moles = 3.55 moles
P = 670/760 atm = 0.88
T = 273+27 = 300 deg K
R = 0.0821
PV=nRT
V=nRT/P
V=3.55 x 0.0821 x 300 / 0.88 = 99.4 L
NH3 Mol.mass = 14 + 3 = 17. = 17g/mole

60.4g ÷ 17g/mole = 3.55 moles

1 mole occupies 22.4 L at STP.
3.55 moles x 22.4 L/mole = 79.5 L
1atm x 79.5 x 300 = 0.9atm x V2 x 273
V2 = (22.4 x 300) ÷ (0.9 x 273)
V2 = 23,856 ÷ 245.7
New volume (V2) = 97.1 Litres.

The answers post by the user, for information only, FunQA.com does not guarantee the right.



More Questions and Answers:
  • Ok my last Sodium acetate question?
  • Does anyone know how to assign H's or C's to an NMR spectra?
  • Balancing equations?
  • What is the molar mass of Copper(II) Sulfate,CuSO4?
  • 9.What mass of BaSO4 will dissolve in 450 mL of aqueous solution. Ksp for BaSO4 is 1 x 10-10.?
  • Freezing Point?
  • Solubility Question... Please Help?
  • What is iodine used for?
  • Can you please classify each of the following compunds as ionic or covalent and why, and give its name?