Question

How many moles of water and oxygen are produced by the decomposition of 86.0grams of hydrogen peroxide?

Answer 1

The balanced equation of the reaction described is:

`2H_2O_2\rightarrow2H_2O+O_2`

We see that two moles of hydrogen peroxide form two moles of water. and two moles of hydrogen form 1 mol of oxygen.

First, we must calculate the moles of peroxide present in 86.0 grams, for which we divide the mass by the molar mass of the peroxide. The molar mass of peroxide is:34.0147g/mol.

The moles of H2O2 will be:

`\begin{gathered} molH_2O_2=givengH_2O_2*(1molH_2O_2)/(MolarMass,gH_2O_2) \\ molH_2O_2=86.0gH_2O_2*(1molH_2O_2)/(34.0147gH_2O_2)=2.53molH_2O_2 \end{gathered}`

Now, the ratio H2O to H2O2 is 2/2, so the moles of water will be:

`\begin{gathered} molH_2O=givenmolH_2O_2*(2molH_2O)/(2molH_2O_2) \\ molH_2O=2.53molH_2O_2*(1molH_2O)/(1molH_2O_2)=2.53molH_2O \end{gathered}`

The ratio O2 to H2o2 is 1/2, so the moles of oxygen will be:

`\begin{gathered} molO_2=givenmolH_2O_2*(1molO_2)/(2molH_2O_2) \\ molO_2=2.53molH_2O_2*(1molO_2)/(2molH_2O_2)=1.26molO_2 \end{gathered}`

Answer: By the decomposition of 86.0 grams of hydrogen peroxide are produced 2.53 moles of water and 1.26 moles of oxygen gas.