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Some of the content of this guide was modeled after a guide originally created by **Openstax**** **and has been adapted for the GPRC Learning Commons in January 2021. This work is licensed under a Creative Commons BY NC SA 4.0 International License.

**Stoichiometry**

In a chemical reaction, stoichiometry is the quantitative description of the proportions of products and reactants. In order to use stoichiometry to determine the quantitative data of either reactants or products, one should be able to **balance chemical reactions** first.

Whether you're aiming to calculate the number of moles, volume, or mass using stochiometry, you can use the following order of steps in your calculation:

**1)** Write a balanced chemical equation.

**2)** Calculate the amount of starting specie using the information given to you in moles. You can use the following equations to find the moles of starting material depending on its type:

**Solids **

n = m/M

**Solutions**

n = CV

**Gases**

n = PV/RT

**3)** Determine if there is a limiting reagent.

**4)** Calculate the number of moles of desired specie by using molar ratio.

**5)** With the number of moles calculated, you can then calculate the desired quantity using the same formula as **step 2 **above.

**Example 1**

How many moles of Na_{2}S_{2}O_{3} is needed to react completely with 52.7 g of AgBr?

**Na _{2}S_{2}O_{3} + AgBr **

**Answer:**

__ Step 1__: Balance the chemical equation:

**2Na _{2}S_{2}O_{3} + AgBr ⟶ NaBr + Na_{3}[Ag(S_{2}O_{3})_{2}]**

__ Step 2__: Calculate the amount of starting specie (AgBr) in moles using the following formula:

**n = m/M**

m (mass) = 52.7g

M (molar mass) = (107.87 g/mol) + (79.90 g/mol) = 187.77 g/mol

__ Step 3__: Determine that the limiting reagent is

__ Step 4__: Calculate the desired amount of

Solving for "X" in the above equation gives:

**X = 0.562 mol**

Therefore, the amount of Na_{2}S_{2}O_{3 } needed is **0.562 mol **to react with 0.281 mol of AgBr.

**Example 2**

Carbon dioxide reacts with lithium hydroxide to produce lithium carbonate and water. What mass of lithium hydroxide is required to react with 1.00 x10^{3} grams of carbon dioxide?

__ Step 1__: Write the balanced equation as shown below:

**CO _{2(g) }+ 2LiOH_{(s) } ⟶ Li_{2}CO_{3(s)} + H_{2}O_{(g)}**

__ Step 2__: Calculate the amount of starting specie (CO

**n = m/M**

m (mass) = **2.00 x10 ^{3}g**

M (molar mass) = (12.01 g/mol) + (16.00 g/mol) = **44.01 g/mol**

= **22.7 mols**

__ Step 3__: Determine the limiting reagent is

__ Step 4__: Calculate the desired amount of

Solving for "X" in the above equation gives:

**X = 45.4 mol**

Therefore, the amount of LiOH needed is **45.4 mol** to react with 22.7 mol of AgBr.

From the mols of LiOH calculated, the mass can be calculated using the following formula:

**n = m/M**

m (mass) = ?

M (molar mass) = (6.94 g/mol) + (16.00 g/mol) +(1.01 g/mol) = 23.95 g/mol

**m = n*M = 22.7 mol * 23.95 g/mol = 543.665 g = 5.44 x10 ^{2}g**

- Last Updated: Sep 21, 2021 2:19 PM
- URL: https://libguides.gprc.ab.ca/chemistry
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