Question

< CH7HW 11/07/19 di 12 out of 23 This is a Mixed question/It is worth 1 point/ You have unlimited attempts / The 16 Question (1 point) The reaction of Cr2O3 with silicon metal at high temperatures will make chromium metal. 2Cr,O3(s) +381(s) — 4Cr(1) + 3Si0,(s) The reaction is begun with 197.00 g of Si and 122.00 g of Cr2O3 1st attempt Part 1 (0.5 point) Select the limiting reactant. < 16/20 > ace → colo 3 4 5 6 7 1

Answer 1

1)

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


mass(Cr2O3)= 122.0 g

use:
number of mol of Cr2O3,
n = mass of Cr2O3/molar mass of Cr2O3
=(1.22*10^2 g)/(1.52*10^2 g/mol)
= 0.8026 mol

Molar mass of Si = 28.09 g/mol


mass(Si)= 197.0 g

use:
number of mol of Si,
n = mass of Si/molar mass of Si
=(1.97*10^2 g)/(28.09 g/mol)
= 7.013 mol
Balanced chemical equation is:
2 Cr2O3 + 3 Si ---> 4 Cr + 3 SiO2


2 mol of Cr2O3 reacts with 3 mol of Si
for 0.8026 mol of Cr2O3, 1.204 mol of Si is required
But we have 7.013 mol of Si

so, Cr2O3 is limiting reagent
Answer: Cr2O3

2)
According to balanced equation
mol of Si reacted = (3/2)* moles of Cr2O3
= (3/2)*0.8026
= 1.204 mol
mol of Si remaining = mol initially present - mol reacted
mol of Si remaining = 7.013 - 1.204
mol of Si remaining = 5.809 mol


Molar mass of Si = 28.09 g/mol

use:
mass of Si,
m = number of mol * molar mass
= 5.809 mol * 28.09 g/mol
= 1.632*10^2 g
Answer: 163.2 g