Question

A) For the following reaction, 4.22 grams of carbon (graphite) are allowed to react with 15.1 grams of oxygen gas. carbon (graphite) (s) + oxygen (g) carbon dioxide (g) What is the maximum amount of carbon dioxide that can be formed? grams What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams

B) For the following reaction, 60.4 grams of iron(III) oxide are allowed to react with 24.7 grams of aluminum. iron(III) oxide (s) + aluminum (s) aluminum oxide (s) + iron (s) What is the maximum amount of aluminum oxide that can be formed? grams What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams

Answer 1

A)

1)
Molar mass of C = 12.01 g/mol


mass(C)= 4.22 g

use:
number of mol of C,
n = mass of C/molar mass of C
=(4.22 g)/(12.01 g/mol)
= 0.3514 mol

Molar mass of O2 = 32 g/mol


mass(O2)= 15.1 g

use:
number of mol of O2,
n = mass of O2/molar mass of O2
=(15.1 g)/(32 g/mol)
= 0.4719 mol
Balanced chemical equation is:
C + O2 ---> CO2


1 mol of C reacts with 1 mol of O2
for 0.3514 mol of C, 0.3514 mol of O2 is required
But we have 0.4719 mol of O2

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


Molar mass of CO2,
MM = 1*MM(C) + 2*MM(O)
= 1*12.01 + 2*16.0
= 44.01 g/mol

According to balanced equation
mol of CO2 formed = (1/1)* moles of C
= (1/1)*0.3514
= 0.3514 mol


use:
mass of CO2 = number of mol * molar mass
= 0.3514*44.01
= 15.46 g
Answer: 15.5 g

2)
C is limiting reagent
Answer: C

3)
According to balanced equation
mol of O2 reacted = (1/1)* moles of C
= (1/1)*0.3514
= 0.3514 mol
mol of O2 remaining = mol initially present - mol reacted
mol of O2 remaining = 0.4719 - 0.3514
mol of O2 remaining = 0.1205 mol


Molar mass of O2 = 32 g/mol

use:
mass of O2,
m = number of mol * molar mass
= 0.1205 mol * 32 g/mol
= 3.856 g
Answer: 3.86 g

B)
1)

Molar mass of Fe2O3,
MM = 2*MM(Fe) + 3*MM(O)
= 2*55.85 + 3*16.0
= 159.7 g/mol


mass(Fe2O3)= 60.4 g

use:
number of mol of Fe2O3,
n = mass of Fe2O3/molar mass of Fe2O3
=(60.4 g)/(1.597*10^2 g/mol)
= 0.3782 mol

Molar mass of Al = 26.98 g/mol


mass(Al)= 24.7 g

use:
number of mol of Al,
n = mass of Al/molar mass of Al
=(24.7 g)/(26.98 g/mol)
= 0.9155 mol
Balanced chemical equation is:
Fe2O3 + 2 Al ---> Al2O3 + 2 Fe


1 mol of Fe2O3 reacts with 2 mol of Al
for 0.3782 mol of Fe2O3, 0.7564 mol of Al is required
But we have 0.9155 mol of Al

so, Fe2O3 is limiting reagent
we will use Fe2O3 in further calculation


Molar mass of Al2O3,
MM = 2*MM(Al) + 3*MM(O)
= 2*26.98 + 3*16.0
= 101.96 g/mol

According to balanced equation
mol of Al2O3 formed = (1/1)* moles of Fe2O3
= (1/1)*0.3782
= 0.3782 mol


use:
mass of Al2O3 = number of mol * molar mass
= 0.3782*1.02*10^2
= 38.56 g
Answer: 38.6 g

2)
Answer: Fe2O3

3)
According to balanced equation
mol of Al reacted = (2/1)* moles of Fe2O3
= (2/1)*0.3782
= 0.7564 mol
mol of Al remaining = mol initially present - mol reacted
mol of Al remaining = 0.9155 - 0.7564
mol of Al remaining = 0.1591 mol


Molar mass of Al = 26.98 g/mol

use:
mass of Al,
m = number of mol * molar mass
= 0.1591 mol * 26.98 g/mol
= 4.292 g
Answer: 4.29 g