Question

Iron (II) Bromide (215.65 g/mol) is reacted with potassium carbonate (138.21 g/mol). If 250.0 grams of both

reactant are used, how much excess reactant will remain after the reaction.
Scratch Work:

Answer 1

Answer: 89.83 g of
`K_2CO_3` will be left as excess reactant after the reaction.

Step-by-step explanation:

To calculate the moles :

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

`\text{Moles of} FeBr_2=(250.0g)/(215.65g/mol)=1.159moles`

`\text{Moles of} K_2CO_3=(250.0g)/(138.21g/mol)=1.809moles`

The balanced chemical reaction is:

`FeBr_2+K_2CO_3\rightarrow 2KBr+FeCO_3`

According to stoichiometry :

1 mole of
`FeBr_2` require = 1 mole of
`K_2CO_3`

Thus 1.159 mole of
`FeBr_2` will require=
`(1)/(1)* 1.159=1.159moles` of
`K_2CO_3`

Thus
`FeBr_2` is the limiting reagent as it limits the formation of product and
`K_2CO_3` is the excess reagent.

moles of
`K_2CO_3` left = (1.809-1.159) = 0.650

Mass of
`K_2CO_3`=
`moles* {\text {Molar mass}}=0.650moles* 138.2g/mol=89.83g`

Thus 89.83 g of
`K_2CO_3` will be left as excess reactant after the reaction.