**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.

F_{2} (g) + NH_{3} (g) –> N_{2}F_{4} (g) + HF (g) [unbalanced]

5.33 mol 2.22 mol

**Answer to Practice Problem #2:**

**1.** **Balance your equation first.**

**5**F_{2} (g) + **2**NH_{3} (g) –> N_{2}F_{4} (g) + **6**HF (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 NH_{3} would need to react with 5.33 moles F_{2}:

* From your balanced equation, the ratio is 2 mol NH_{3} : 5 mol F_{2}.

5.33 mol F_{2} X ( 2 mol NH_{3} / 5 mol F_{2} ) = 2.132 mol NH_{3} needed

* But you have 2.22 mol NH_{3} to start with and need 2.132 mol NH_{3 } to react, so you have extra NH_{3} and **limiting reagent is F _{2}**

* To calculate excess: 2.220 – 2.132 = 0.088 mol NH_{3} in excess

then convert to grams and get **1.496 g NH _{3} in excess**

OR

Figure out how many moles of F_{2} would need to react with 2.22 moles NH_{3}:

* From your balanced equation, the ratio is 5 mol F_{2} : 2 mol NH_{3}.

2.22 mol NH_{3} X ( 5 mol F_{2} / 2 mol NH_{3} ) = 5.55 mol F_{2} needed

* But you have 5.33 mol F_{2} to start with and need 5.55 mol F_{2} to react, so you have F_{2} as the limiting reagent because you don’t have enough F_{2} needed to react

* Then to find out how much NH_{3} 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 F_{2} is the limiting reagent, you use the 5.33 moles of F_{2} 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 F_{2} X ( 6 mol HF / 5 mol F_{2} ) = 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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