Question

If 0.700 moles Ag is reacted with 10.0 g S, is sulfur or aluminum the limiting reactant?

Answer 1

Answer:
`S` is the limiting reagent

Step-by-step explanation:

To calculate the moles :

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

`\text{Moles of} S=(10.0g)/(32g/mol)=0.3125moles`

`\text{Moles of} Ag=0.700moles`

`2Ag+S\rightarrow Ag_2S`

According to stoichiometry :

1 mole of
`S` require = 2 moles of
`Ag`

Thus 0.3125 moles of
`S` will require=
`(2)/(1)* 0.3125=0.6250moles` of
`Ag`

Thus
`S` is the limiting reagent as it limits the formation of product and
`Ag` is the excess reagent.