Answer:
O2 will be the limiting reactant while CS2 will be the excess reactant
Explanations:
Given the reaction between CS2 and oxygen expressed as:
[tex]CS_2+3O_2\rightarrow CO_2+2SO_2[/tex]
Given the following pramters
Mass of CS2 = 20grams
Mass of O2 = 10grams
Determine the mole of CS2
[tex]\begin{gathered} mole=\frac{mass}{molar\text{ mass}} \\ mole\text{ of CS}_2=\frac{20}{76.139} \\ mole\text{ of CS}_2=0.2627mole \end{gathered}[/tex]
Determine the mole of oxygen
[tex]\begin{gathered} mole\text{ of O}_2=\frac{10}{2(16)} \\ mole\text{ of O}_2=\frac{10}{32}=0.3125moles \\ \end{gathered}[/tex]
Since there are 3 atoms of oxygen, 1 atom will have:
[tex]1\text{ atom of }O_2=\frac{0.3125}{3}=0.1042mole[/tex]
Since the mole of oxygen is less than the mole ofCS2, hence O2 will b the limiting reactant while CS2 will be the excess reactant
