**The key steps (in order) to answering a stoichiometry problem:**

1. Balance your equation first.

2. Convert from grams to moles using molar mass.

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

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

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

6. Calculate percent yield [if necessary] (Divide actual yield by theoretical yield X 100%)

**Sample Example Problem [to show you how it’s done]:**

**Problem:** How many grams of silicon carbide are produced when 50.0 g of silicon dioxide is heated with 32.0 g of carbon?

SiO_{2} (s) + C (s) –> SiC (s) + CO (g) [unbalanced]

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

SiO_{2} (s) + **3**C (s) –> SiC (s) + **2**CO (g) [balanced]

**2.** **Convert from grams to moles using molar mass.**

To get moles from grams of silicon dioxide (SiO_{2}): [molar mass = 60 g/mol]

50.0 g SiO_{2} X ( 1 mol / 60 g ) = 0.833 mol SiO_{2}

To get moles from grams of carbon (C): [molar mass = 12 g/mol]

32.0 g C X ( 1 mol / 12 g ) = 2.667 mol C

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

Figure out how many moles of C would need to react with 0.833 moles SiO_{2}:

* From your balanced equation, the ratio is 3 mol C : 1 mol SiO_{2}.

0.833 mol SiO_{2} X ( 3 mol C / 1 mol SiO_{2} ) = 2.5 mol C needed

* But you have 2.667 mol C to start with and need 2.5 mol C to react, so you have extra C and limiting reagent is SiO_{2}

* To calculate excess: 2.667 – 2.500 = 0.167 mol C in excess

OR

Figure out how many moles of SiO_{2} would need to react with 0.833 moles C:

* From your balanced equation, the ratio is 1 mol SiO_{2} : 3 mol C.

2.667 mol C X ( 1 mol SiO_{2} / 3 mol C ) = 0.889 mol SiO_{2} needed

* But you have 0.833 mol SiO_{2} to start with and need 0.889 mol SiO_{2} to react, so you have SiO_{2} limiting reagent because you don’t have enough SiO_{2} needed to react

* Compare 0.889 mol SiO_{2} needed vs. 0.833 mol SiO_{2} you started with

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

Since SiO_{2} is the limiting reagent, you use the moles of SiO_{2} if you have to calculate how many grams of silicon carbide (the product you’re trying to find) are being produced. Use your mole ratios from your balanced equation:

0.833 mol SiO_{2} X ( 1 mol SiC / 1 mol SiO_{2} ) = 0.833 mol SiC produced

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

To get grams from moles of carbon (SiC): [molar mass = 40 g/mol]

0.833 mol SiC X ( 40 g / 1 mol ) = **33.33 g SiC produced**

**Answer to Problem:** **33.33 g SiC produced = the theoretical yield**

**Practice Problems Here:**

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