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Chapter 9: Problem 73

Consider the balanced chemical equation \(2 \mathrm{H}_{2} \mathrm{O}_{2} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}\) (a) Given \(20.0 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}_{2}\) (hydrogen peroxide), how many grams of water will the reaction yield? (b) How many grams of hydrogen peroxide would you need to get \(20.0 \mathrm{~g}\) of water? (c) How many grams of hydrogen peroxide would you need to get \(20.0 \mathrm{~g}\) of \(\mathrm{O}_{2} ?\)

Short Answer

Expert verified
(a) The reaction will yield \(10.6 \, g\) of water. (b) You would need \(37.7 \, g\) of hydrogen peroxide to get \(20 \, g\) of water. (c) You would need \(42.5 \, g\) of hydrogen peroxide to get \(20 \, g\) of \(\mathrm{O}_{2}\).

Step by step solution

01

Write down the balanced chemical equation

Given the balanced chemical equation: \[2 \mathrm{H}_{2} \mathrm{O}_{2} \rightarrow 2 \mathrm{H}_{2} \mathrm{O} + \mathrm{O}_{2}\]
02

Calculate the molar masses of the substances involved

We will need to convert between moles and grams using the molar masses of the reactants and products. The molar mass of \(\mathrm{H}_{2} \mathrm{O}_{2}\) is approximately \(34.0 \, g/mol\), of \(\mathrm{H}_{2} \mathrm{O}\) is approximately \(18.0 \, g/mol\), and of \(\mathrm{O}_{2}\) is approximately \(32.0 \, g/mol\). (a)
03

Step 3a: Convert given mass of hydrogen peroxide to moles

We are given \(20.0 \, g\) of \(\mathrm{H}_{2} \mathrm{O}_{2}\). To convert to moles, we use the molar mass of hydrogen peroxide: \[\frac{20.0 \, g}{34.0 \, g/mol} = 0.588 \, mol\]
04

Step 4a: Use mole-to-mole ratio to find moles of water produced

From the balanced equation, we know that \(2 \, mol\) of \(\mathrm{H}_{2} \mathrm{O}_{2}\) yield \(2 \, mol\) of \(\mathrm{H}_{2} \mathrm{O}\). So, the moles of water produced are: \[0.588 \, mol \times \frac{2 \, mol \, \mathrm{H}_{2} \mathrm{O}}{2 \, mol \, \mathrm{H}_{2} \mathrm{O}_{2}} = 0.588 \, mol\]
05

Step 5a: Convert moles of water to grams

Finally, we will convert the moles of water produced to grams using the molar mass of water: \[0.588 \, mol \times 18.0 \, \frac{g}{mol} = 10.6 \, g\] So, the reaction will yield \(10.6 \, g\) of water. (b)
06

Step 3b: Convert given mass of water to moles

We are given \(20.0 \, g\) of \(\mathrm{H}_{2} \mathrm{O}\). To convert to moles, we use the molar mass of water: \[\frac{20.0 \, g}{18.0 \, g/mol} = 1.11 \, mol\]
07

Step 4b: Use mole-to-mole ratio to find moles of hydrogen peroxide needed

From the balanced equation, we know that \(2 \, mol\) of \(\mathrm{H}_{2} \mathrm{O}\) are produced by \(2 \, mol\) of \(\mathrm{H}_{2} \mathrm{O}_{2}\). So, the moles of hydrogen peroxide needed are: \[1.11 \, mol \times \frac{2 \, mol \, \mathrm{H}_{2} \mathrm{O}_{2}}{2 \, mol \, \mathrm{H}_{2} \mathrm{O}} = 1.11 \, mol\]
08

Step 5b: Convert moles of hydrogen peroxide to grams

Finally, we will convert the moles of hydrogen peroxide needed to grams using its molar mass: \[1.11 \, mol \times 34.0 \, \frac{g}{mol} = 37.7 \, g\] So, you would need \(37.7 \, g\) of hydrogen peroxide to get \(20 \, g\) of water. (c)
09

Step 3c: Convert given mass of oxygen to moles

We are given \(20.0 \, g\) of \(\mathrm{O}_{2}\). To convert to moles, we use the molar mass of oxygen: \[\frac{20.0 \, g}{32.0 \, g/mol} = 0.625 \, mol\]
10

Step 4c: Use mole-to-mole ratio to find moles of hydrogen peroxide needed

From the balanced equation, we know that one mole of \(\mathrm{O}_{2}\) is produced by \(2 \, mol\) of \(\mathrm{H}_{2} \mathrm{O}_{2}\). So, the moles of hydrogen peroxide needed are: \[0.625 \, mol \times \frac{2 \, mol \,\mathrm{H}_{2} \mathrm{O}_{2}}{1 \, mol \, \mathrm{O}_{2}} = 1.25 \, mol\]
11

Step 5c: Convert moles of hydrogen peroxide to grams

Finally, we will convert the moles of hydrogen peroxide needed to grams using its molar mass: \[1.25 \, mol \times 34.0 \, \frac{g}{mol} = 42.5 \, g\] So, you would need \(42.5 \, g\) of hydrogen peroxide to get \(20 \, g\) of \(\mathrm{O}_{2}\).

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