Question

I need to produce 500 g of lithium oxide(Li2O)a) how many grams of Lithium and b) how many liters of oxygen do I need

Answer 1

The balanced reaction to produce lithium oxide is:

`4Li+O_2\rightarrow2Li_2O`

To produce 2 moles of lithium oxide, 4 moles of lithium and 1 mole of oxygen are needed.

The first step to solve these questions is to convert the mass of lithium to moles using its molecular weight (MW=29.88g/mol):

`500g\cdot(molLi_2O)/(29.88g)=16.73molLi_2O`

a. Use the ratio of moles of lithium to moles of lithium oxide to find how many moles of lithium are needed to produce 16.73moles of lithium oxide:

`16.73molLi_2O\cdot(4molLi)/(2molLi_2O)=33.46molLi`

Convert this to mass using lithium molecular weight:

`33.46molLi\cdot(6.94g)/(molLi)=232.21g`

232.21 grams of Lithium are needed.

b. Follow the same procedure to find the moles of O2 needed:

`16.73molLi_2O\cdot(1molO_2)/(2molLi_2O)=8.365molO_2`

At this point we can use the ideal gas law to find the volume occupied by 8.365 moles of O2. Use STP (P=1atm, T=273.15K) and R as 0.082atmL/molK:

`\begin{gathered} PV=nRT \\ V=(nRT)/(P) \\ V=(8.365mol\cdot0.082atmL/molK\cdot273.15K)/(1atm) \\ V=187.36 \end{gathered}`

It means that 187.36L of O2 are needed.