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Lowell High School Chemistry

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Practice Problem #2:

Identify the limiting reagent for the given combination of reactants. Calculate how many grams of whatever substance will be in excess and how many grams of HF will be made.

F2 (g)  +  NH3 (g)  –>  N2F4 (g)  +  HF (g)               [unbalanced]

5.33 mol    2.22 mol

5F2 (g)  +  2NH3 (g)  –>  N2F4 (g)  +  6HF (g)                                 [balanced]

2. Convert from grams to moles using molar mass.

* Not necessary because reactant amounts are given in moles already.

3. Determine the limiting reagent [if necessary] (Use mole ratios to figure out.)

Figure out how many moles of NH3 would need to react with 5.33 moles F2:

* From your balanced equation, the ratio is 2 mol NH3 : 5 mol F2.

5.33 mol F2  X  ( 2 mol NH3 / 5 mol F2 )  = 2.132 mol NH3 needed

* But you have 2.22 mol NH3 to start with and need 2.132 mol NH to react, so you have extra NH3 and limiting reagent is F2

* To calculate excess: 2.220 – 2.132 = 0.088 mol NH3 in excess

then convert to grams and get 1.496 g NH3 in excess

OR

Figure out how many moles of F2 would need to react with 2.22 moles NH3:

* From your balanced equation, the ratio is 5 mol F2 : 2 mol NH3.

2.22 mol NH3  X  ( 5 mol F2 / 2 mol NH3 )  = 5.55 mol F2 needed

* But you have 5.33 mol F2 to start with and need 5.55 mol F2 to react, so you have F2 as the limiting reagent because you don’t have enough F2 needed to react

* Then to find out how much NH3 is in excess, do the part directly preceding this (part 3a).

4. Use ratios to find the moles of the reactant or product you need to find.

Since F2 is the limiting reagent, you use the 5.33 moles of F2 if you have to calculate how many moles of HF (the product you’re trying to find) are being produced. Use your mole ratios from your balanced equation:

5.33 mol F2  X  ( 6 mol HF / 5 mol F2 )  = 6.396 mol HF produced

5. Convert from moles back to grams of the new substance using that substance’s molar mass.

To get grams from moles of HF:         [molar mass = 20 g/mol]

6.396 mol HF  X  ( 20 g  /  1 mol ) = 127.92 g HF produced

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