**Lowell High School Chemistry**

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**Practice Problem #4:**

Given the following reaction, answer Parts A and B.

NaClO_{3} (s) –> NaCl (s) + O_{2} (g) [unbalanced]

Part A: 12.00 moles of NaClO_{3} will produce how many grams of O_{2}?

Part B: How many grams of NaCl are produced when 80.0 grams of O_{2} are produced?

**Answer to Practice Problem #4 (Part A):**

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

**2**NaClO_{3} (s) –> **2**NaCl (s) + **3**O_{2} (g) [balanced]

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

12.00 moles of NaClO_{3} is already given so you don’t need to convert to moles.

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

NaClO_{3} is the only reactant and therefore the limiting reagent in this reaction.

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

What you are trying to find is grams of O_{2}, so you need to find the moles of O_{2} first. The mole ratio comparing O_{2} to NaClO_{3} (as given in your balanced equation) is 3 mol O_{2} : 2 mol NaClO_{3}.

12.00 mol NaClO_{3} X ( 3 mol O_{2 } / 2 mol NaClO_{3} ) = 18.00 mol O_{2} produced

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

To get grams from moles of O_{2}: [molar mass = 32 g/mol]

18.00 mol O_{2 } X ( 32 g / 1 mol ) = **576.00 g O _{2}**

**produced**

**Answer to Problem #4 (Part A):** **576.00 g O _{2}**

**produced**

**Answer to Practice Problem #4 (Part B):**

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

**2**NaClO_{3} (s) –> **2**NaCl (s) + **3**O_{2} (g) [balanced]

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

To get grams from moles of O_{2}: [molar mass = 32 g/mol]

80.00 g O_{2 } X ( 1 mol / 32 g ) = 2.50 mol O_{2} produced

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

Not necessary.

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

What you are trying to find is grams of NaCl produced, so you need to find the moles of NaCl first. The mole ratio comparing NaCl to O_{2} (as given in your balanced equation) is 2 mol NaCl : 3 mol O_{2}.

2.50 mol O_{2} X ( 2 mol NaCl _{ }/ 3 mol O_{2} ) = 1.666667 mol NaCl produced

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

To get grams from moles of NaCl: [molar mass = 58.45 g/mol]

1.666667 mol NaCl X ( 58.45 g / 1 mol ) = **97.417 g NaCl produced**

**Answer to Problem #4 (Part B):** **97.417 g NaCl produced**

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