Question

if the percent yield for the following reaction is 65.0%, how many grams of kclo3 are needed to produce 4.00 g of o2?

Answer 1

Balanced Equation:

The decomposition of potassium chlorate (KClO₃) produces oxygen gas (O₂):

2KClO₃→2KCl+3O₂

​Given:

Desired yield of O₂= 4.00 g

Percent yield = 65.0%

Step 1: Calculate Theoretical Yield of O₂ :

The theoretical yield is the amount of product that should be obtained in an ideal scenario.

From the balanced equation:

2 moles of KClO₃ produce 3 moles of O₂.

Calculate the molar mass of O2: 32.00g/mol (Oxygen)

Use the molar mass to convert grams of O₂ to moles:

Moles of O₂=Mass/Molar mass=4.00 g/32.00g/mol=0.125mol

Calculate Theoretical Yield:

From the balanced equation, 2 moles of KClO₃ produce 3 moles of O₂.

If 3 moles of O₂ require 2 moles of KClO₃, then 0.125 moles of O₂ require:

Moles of KClO₃=0.125mol O₂×2 mol KClO₃/3 mol O₂=0.0833 mol KClO₃

​Step 2: Calculate Mass of KClO₃ Needed (with Percent Yield):

The actual yield is 65.0% of the theoretical yield.

Calculate Theoretical Yield:

Theoretical Yield=Moles of KClO₃×Molar mass of KClO₃

Theoretical Yield=0.0833mol×Molar mass of KClO₃

​Calculate Mass of KClO₃ (with Percent Yield):

Percent Yield=Actual Yield/Theoretical Yield×100

Given the Percent Yield = 65.0%, rearrange the equation to find the Actual Yield.

Rearranged equation for Actual Yield:

Actual Yield=Percent Yield/100×Theoretical Yield

Substitute the given values to find the Actual Yield.

Calculate Actual Mass of KClO₃:

Actual Mass of KClO₃=Actual Yield=65.0/100×Theoretical Yield

This calculation will yield the mass of KClO₃ needed to produce 4.00 grams of O₂ with a percent yield of 65.0%.

Answer 2

Approximately 10.21 grams of KClO₃ are needed to produce 4.00 grams of O₂ with a percent yield of 65.0%.

To determine the amount of KClO₃ needed to produce a given amount of O₂ with a known percent yield, we'll use the balanced chemical equation for the reaction and the concept of percent yield.

Let's assume the reaction is represented by the following balanced equation:

`\[ 2 \, \text{KClO}_3 \rightarrow 2 \, \text{KCl} + 3 \, \text{O}_2 \]`

This equation indicates that 2 moles of KClO₃ produce 3 moles of O₂.

1. Convert the given mass of O₂ to moles using the molar mass of O₂:

`\[ \text{Molar mass of O}_2 = 2 * \text{Atomic mass of O} \]`

`\[ \text{Molar mass of O}_2 = 2 * 16.00 \, \text{g/mol} = 32.00 \, \text{g/mol} \]`

`\[ \text{Moles of O}_2 = \frac{\text{Given mass}}{\text{Molar mass}} \]`

`\[ \text{Moles of O}_2 = \frac{4.00 \, \text{g}}{32.00 \, \text{g/mol}} = 0.125 \, \text{mol} \]`

2. Use the stoichiometry of the reaction to find the moles of KClO₃ needed:

- From the balanced equation, 2 moles of KClO₃ produce 3 moles of O₂.

`\[ \text{Moles of KClO}_3 = \frac{\text{Moles of O}_2}{3} * 2 \]`

`\[ \text{Moles of KClO}_3 = \frac{0.125 \, \text{mol}}{3} * 2 \]`

`\[ \text{Moles of KClO}_3 = 0.0833 \, \text{mol} \]`

3. Calculate the mass of KClO₃ needed using its molar mass:

`\[ \text{Molar mass of KClO}_3 = \text{Atomic mass of K} + \text{Atomic mass of Cl} + 3 * \text{Atomic mass of O} \]`

`\[ \text{Molar mass of KClO}_3 = 39.10 + 35.45 + 3 * 16.00 \, \text{g/mol} \]`

`\[ \text{Molar mass of KClO}_3 = 122.55 \, \text{g/mol} \]`

`\[ \text{Mass of KClO}_3 = \text{Moles of KClO}_3 * \text{Molar mass of KClO}_3 \]`

`\[ \text{Mass of KClO}_3 = 0.0833 \, \text{mol} * 122.55 \, \text{g/mol} \]`

`\[ \text{Mass of KClO}_3 = 10.21 \, \text{g} \]`

So, approximately 10.21 grams of KClO₃ are needed to produce 4.00 grams of O₂ with a percent yield of 65.0%. The percent yield accounts for the efficiency of the reaction, and the actual yield would be 65.0% of the theoretical yield.