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How to do gas calculations and what out what gas and how much of that gas is produced?

The question is:

When 100cm3 of hydrogen bromide reacts with 80cm3 of ammonia, a white solid is for,ed and some gas is left over. What gas and how much of it is left over?

NH3(g) + HBr(g) --> NH4Br(s)

1 Answer

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  • Dr W
    Lv 7
    9 years ago
    Favorite Answer

    as in ALL stoichiometry problems.. (how much of this or that reacts or is formed or what is yield or % yield.. etc), follow these steps!

    (1).. write a balanced equation

    (2).. convert everything given to moles

    (3).. determine limiting reagents

    (4).. convert moles limiting reagents to moles other species using the coefficients of the balanced equation...

    (5).. convert moles back to mass.. this is theoretical mass.. aka theoretical yield

    (6).. % yield = (actual mass recovered / theoretical mass) x 100%

    the idea being the coefficients of the balanced equation are in mole ratios. So if we write a balanced equation first, then convert everything to moles, we can use those coefficients to convert between the different chemical species.

    ******

    so.. 1 thing to note... in the special case of gases at constant T and P... mole ratios = volume ratios...

    i.e..

    the coefficients in...

    1 NH3(g) + 1 HBr(g) ---> 1 NH4Br(s)

    can be read as...

    1 mole NH3 molecules reacts with 1 mole HBr molecules to form 1 mole NH4Br molecules

    AND...

    1 L of NH3 reacts wtih 1 L of HBr

    why can we do that?

    PV = nRT

    V/n = RT/P = a constant.. at constant T and P

    so..

    V1/n1 = V2/n2

    V1/V2 = n1/n2... and volume ratios = mole ratios... capice?

    i.e. we can work in cm³ instead of converting to moles.

    now that we have that out of the way..

    *** 1 ***

    1 NH3(g) + 1 HBr(g) --> 1 NH4Br(s).. is balanced.. always double check!

    *** 2 ***

    again.. we'll work in cm³ instead of moles...

    we have 100 cm³ HBr

    we have 80 cm³ NH3

    so we don't need to convert anything

    *** 3 ***

    from the balanced equation 1 cm³ NH3 reacts with 1 cm³ HBr

    from (2).. we know we have 80cm³ NH3.. so we NEED...

    80 cm³ NH3 x (1 cm³ HBr / 1 cm³ NH3) = 80 cm³ HBr

    and since HAVE 100 cm³ HBr available, and only need 80 cm³ to completely consume all the NH3, we have 20 cm³ "HBr left over... and HBr is in XS (eXcesS) and NH3 is the limiting reagent

    ***********

    and that's where we stop...

    20 cm³ of HBr is left over

    *******

    how much of this did you follow?

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