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 80.00 g of Si and 147.00 g of Cr2O3.

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

Answer 1

Molar mass of Cr2O3,
MM = 2*MM(Cr) + 3*MM(O)
= 2*52.0 + 3*16.0
= 152 g/mol


mass(Cr2O3)= 147.0 g

use:
number of mol of Cr2O3,
n = mass of Cr2O3/molar mass of Cr2O3
=(147.0 g)/(152 g/mol)
= 0.9671 mol

Molar mass of Si = 28.09 g/mol


mass(Si)= 80.0 g

use:
number of mol of Si,
n = mass of Si/molar mass of Si
=(80.0 g)/(28.09 g/mol)
= 2.848 mol
Balanced chemical equation is:
2 Cr2O3 + 3 Si ---> 4 Cr + 3 SiO2


2 mol of Cr2O3 reacts with 3 mol of Si
for 0.9671 mol of Cr2O3, 1.451 mol of Si is required
But we have 2.848 mol of Si

so, Cr2O3 is limiting reagent
we will use Cr2O3 in further calculation

According to balanced equation
mol of Si reacted = (3/2)* moles of Cr2O3
= (3/2)*0.9671
= 1.451 mol
mol of Si remaining = mol initially present - mol reacted
mol of Si remaining = 2.848 - 1.451
mol of Si remaining = 1.397 mol


Molar mass of Si = 28.09 g/mol

use:
mass of Si,
m = number of mol * molar mass
= 1.397 mol * 28.09 g/mol
= 39.25 g
Answer: 39.2 g