What volume of gas is needed to prepare a 20% NH3 solution?

An aqueous solution of NH3 which is 20.00% NH3 by mass has a density of 0.9250 g/ml. What volume of NH3 gas at STP would be required in the preparation of 2.500 L of solution?
Catherine

This problem is can be solved by breaking it into a chain of simpler problems. Follow these steps.
  1. Identify the unknown, including units. In your problem, the unknown is the volume of NH3 gas. Units aren't specified, so we're free to choose any units we like. We'll try liters of NH3 gas.

    ? L NH3 gas

  2. List the given information. You know the following:
    • The aqueous solution of NH3 is 20.00% NH3 by mass, which means that there are 20.00 grams of NH3 per 100 grams of solution (20.00 g NH3 = 100 g solution).
    • The solution has a density of 0.9250 g/mL, which means that 0.9250 g solution = 1 mL solution.
    • 2.500 L of solution is being prepared.
    • The gas is at STP, meaning it has a temperature of 273.15 K and a pressure of 1 atm.
  3. Connect the given information with the unknown. Try to work backwards from the unknown. The only information given about the gaseous NH3 is that the gas is at STP. Assume that the gas is ideal. Then you can connect L NH3 gas with moles of NH3, using the temperature (273 K) and the pressure (1 atm) and the ideal gas law. (You can make the same connection using the fact that 1 mole of ideal gas at STP occupies 22.4 L):

    ?
    mol NH3 1 mol NH3
    = 22.4 L NH3

    L NH3

    The target is now moles of NH3. If you had g NH3, you could easily get moles NH3 using the molecular weight.

    ?
    g NH3 17.0 g NH3 = 1 mol NH3
    mol NH3 1 mol NH3
    = 22.4 L NH3

    L NH3

    The target is now g of NH3. You know that 20.00 g NH3 = 100 g solution, so

    ?
    g solution 100 g soln = 20.00 g NH3
    g NH3 17.0 g NH3 = 1 mol NH3
    mol NH3 1 mol NH3
    = 22.4 L NH3

    L NH3

    The target is now g of solution. From the given density, 0.9250 g solution = 1 mL solution, and

    ?
    mL soln 0.9250 g soln = 1 mL soln
    g solution 100 g soln = 20.00 g NH3
    g NH3 17.0 g NH3 = 1 mol NH3
    mol NH3 1 mol NH3
    = 22.4 L NH3

    L NH3

    Finally, you know that there are 2.500 L solution, so you can easily get mL of solution:

    2.500 L 1000 mL = 1 L
    mL soln 0.9250 g soln = 1 mL soln
    g solution 100 g soln = 20.00 g NH3
    g NH3 17.0 g NH3 = 1 mol NH3
    mol NH3 1 mol NH3
    = 22.4 L NH3

    L NH3

  4. Do the math. Use the roadmap above to set up a string of conversion factors so that all units but L NH3 gas cancel.
  5. Check the result Does the answer make sense? Check to see if you can work back towards the given information from your final answer, or try to devise another way to solve the problem.

Author: Fred Senese senese@antoine.frostburg.edu



General Chemistry Online! What volume of gas is needed to prepare a 20% NH_3_ solution?

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Last Revised 08/17/15.URL: http://antoine.frostburg.edu/chem/senese/101/gases/faq/print-preparing-gas-solutions.shtml