Anonymous asked in Science & MathematicsChemistry · 10 years ago

I need help with this STP chemistry problem?

11.2 L of He gas at STP would contain how many grams?

Please explain how you got the answer.


Where did you get 1 mol He equals 4.00 mol g?

4 Answers

  • 10 years ago
    Favorite Answer

    You need two things: the meaning of STP which is Standard Temperature and Pressure which is 1 atm and 0 C.

    Please note that tables of thermodynamic data are expressed under Standard Conditions, which is 1 atm and 25 C. So do not get confused.

    Now He is certainly an idea gas, so it obeys the ideal gas law. We could work it out, but recall that 1 mole of an idea gas at STP occupies a volume of 22.4 l.

    CHeck PV=nRT V = nRT/P = 1 mole x 0.082057 l atm K-1 mol -1) x 273 K/1 atm = 22.4 l

    But we have only 11.2 l, so half as much. Therefore we have half a mole of He. Since the MW of He is 4, we have 2 grams of He gas.

  • 10 years ago

    At STP, one mole of any gas occupies 22.4 liters.

    So, 11.2 L of He at STP will have this number of moles:

    11.2 / 22.4 = 0.5 mol He which is equivalent to:

    0.5 mol He x 4.0026 g/mol = 2.0 g He

  • 10 years ago

    At STP 1 mole of any gas = 22.4 L.

    With 11.2 L, you have 0.500 of He.

    1.00 mole of He = 4.00 g

    0.500 mole of He = 2.00 g

  • nat
    Lv 4
    4 years ago

    at the beginning, the mole ratio of Mg to HCl is a million:2. From the quantities that's obvious that HCl is the proscribing reactant, i.e. HCl runs out first (using fact as a fashion to react with 0.288 of Mg, you like 2x0.288 = 0.576mol of HCl yet you basically have 0.450 mole HCl!), and as quickly because it runs our the reaction stops (using fact there will be not greater HCl for the the rest Mg (unreacted) to react with!). Now be conscious that for each 2mol of HCl the reaction supplies upward push to 1mol of H2 gas in accordance to the stoichiometry of the reaction. So for 0.450 mol HCl (proscribing reactant) you're able to acquire 0.450/2 mol of H2 gas. We additionally be attentive to that 1mol of an suitable gas has a quantity of twenty-two.4 L at STP. So assuming the suitable habit for the H2(g) generated, its quantity is = (0.450/2) x 22.4L (at STP)

Still have questions? Get your answers by asking now.