Question

The reaction of Cr2O3 with silicon metal at high temperatures will make chromium metal.

2Cr2​O3​(s)+3Si(s) -------> 4Cr(l)+3SiO2(s)

The reaction is begun with 161.00 g of Si and 139.00 g of Cr2O3.

How many grams of the excess reactant is left after the reaction is complete?

Answer 1

Answer: 122.9 g of $Si$ will be left from the given masses of both reactants.

Explanation:

$2Cr_2O_3(s)+3Si(s)\rightarrow 4Cr(l)+3SiO_2(s)$

To calculate the moles :

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

$\text{Moles of} Cr_2O_3=\frac{139.00}{152}=0.91moles$

$\text{Moles of} Si=\frac{161.00}{28}=5.75moles$

According to stoichiometry :

2 moles of $Cr_2O_3$ require 3 moles of $Si$

Thus 0.91 moles of $Cr_2O_3$ will require=$\frac{3}{2}\times 0.91=1.36moles$ of $Si$

Thus $Cr_2O_3$ is the limiting reagent as it limits the formation of product and $Si$ is the excess reagent.

Thus (5.75-1.36) = 4.39 moles  of excess reagent.

Mass of $Si=moles\times {\text {Molar mass}}=4.39moles\times 28g/mol=122.9g$

Thus 122.9 g of $Si$ will be left from the given masses of both reactants.