4 C3H5(NO3)3(s) 12 CO2(g) + 10 H2O(g) + 6 N2(g) + O2(g)?

Calculate the total volume of gases when collected at 1.2 atm and 25°C from 2.9 multiplied by 102 g of nitroglycerin.

im not understanding how to get the number of moles for each ..

Comments

  • am i interpreting this 2.9 multiplied by 102 g

    corrctly as 290. grams

    that's a lot of nitro

    find moles of nitro, using molar mass:

    2.9 e2 grams @ 227.09 g/mol =

    1.277 moles of nitro glycerin

    by the equation

    4 C3H5(NO3)3(s) --> 12 CO2(g) + 10 H2O(g) + 6 N2(g) + O2(g)

    4 moles of nitro produces 29 moles of gases

    so

    1.277 moles of nitro glycerin @ 29 mol gases / 4 mol nitro = 9.258 moles of gases

    PV = nRT

    (1.2 atm)(Litres) = (9.258 moles)(0.08206 L-atm/mol-K)(298K)

    Litres = 188.67 litres of gases

    in your 2 sig fig problem,

    your answer should be rounded off to

    190 Litres of gases

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