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.

Explanation:

To calculate the moles :

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

$\text{Moles of} FeBr_2=\frac{250.0g}{215.65g/mol}=1.159moles$

$\text{Moles of} K_2CO_3=\frac{250.0g}{138.21g/mol}=1.809moles$

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=$\frac{1}{1}\times 1.159=1.159moles$  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 $K_2CO_3$= $moles\times {\text {Molar mass}}=0.650moles\times 138.2g/mol=89.83g$

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