Question

If 20.5 g of chlorine is reacted with 20.5 g of sodium, which reactant is in excess?

Answer 1

Answer:
`Na` is in excess.

Step-by-step explanation:

To calculate the moles :

`\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}`

`\text{Moles of} Cl_2=(20.5g)/(71g/mol)=0.29mol`

`\text{Moles of} Na=(20.5g)/(23g/mol)=0.89moles`

`2Na(s)+Cl_2(g)\rightarrow 2NaCl(s)`

According to stoichiometry :

1 mole of
`Cl_2` require 2 moles of
`Na`

Thus 0.29 moles of
`Cl_2` will require=
`(2)/(1)* 0.29=0.58moles` of
`Na`

Thus
`Cl_2` is the limiting reagent as it limits the formation of product and
`Na` is the excess reagent.