# Question #5f09b

May 7, 2017

WARNING! Long answer! The mass of $\text{Al}$ to be used is 258 g.

#### Explanation:

The $\text{Al}$ will react with the energizer and with every metal oxide in the mixture.

Mass of each reactant

${\text{Mass of Cr"_2"O"_3 = 800 color(red)(cancel(color(black)("g ore"))) × ("55,09 g Cr"_2"O"_3)/(100 color(red)(cancel(color(black)("g ore")))) = "440,7 g Cr"_2"O}}_{3}$

$\text{Mass of FeO" = 800 color(red)(cancel(color(black)("g ore"))) × "16,79 g FeO"/(100 color(red)(cancel(color(black)("g ore")))) = "134,3 g FeO}$

$\text{Mass of CrO"_3 = "60 g}$

$\text{Mass of KClO"_3 = "70 g}$

Mass of $\text{Al}$ consumed by each reactant

(a) By $\text{Cr"_2"O"_3}$

${\text{Cr"_2"O"_3 + "2Al" → "2Cr" + "Al"_2"O}}_{3}$

$\text{Mass of Al" = "440,7" color(red)(cancel(color(black)("g Cr"_2"O"_3))) × (1 color(red)(cancel(color(black)("mol Cr"_2"O"_3))))/("151,99" color(red)(cancel(color(black)("g Cr"_2"O"_3)))) × (2 color(red)(cancel(color(black)("mol Al"))))/(1 color(red)(cancel(color(black)("mol Cr"_2"O"_3)))) × "26,98 g Al"/(1 color(red)(cancel(color(black)("mol Al")))) = "156,5 g Al}$

(b) By $\text{FeO}$

${\text{3FeO + 2Al" → "3Fe" + "Al"_2"O}}_{3}$

$\text{Mass of Al" = "134,3" color(red)(cancel(color(black)("g FeO"))) × (1 color(red)(cancel(color(black)("mol FeO"))))/("71,84" color(red)(cancel(color(black)("g FeO")))) × (2 color(red)(cancel(color(black)("mol Al"))))/(3 color(red)(cancel(color(black)("mol FeO")))) × "26,98 g Al"/(1 color(red)(cancel(color(black)("mol Al")))) = "33,62 g Al}$

(c) By ${\text{CrO}}_{3}$

${\text{CrO"_3 + "2Al" → "Cr" + "Al"_2"O}}_{3}$

$\text{Mass of Al" = 60 color(red)(cancel(color(black)("g CrO"_3))) × (1 color(red)(cancel(color(black)("mol CrO"_3))))/("99,99" color(red)(cancel(color(black)("g CrO"_3)))) × (2 color(red)(cancel(color(black)("mol Al"))))/(1 color(red)(cancel(color(black)("mol CrO"_3)))) × "26,98 g Al"/(1 color(red)(cancel(color(black)("mol Al")))) = "32,4 g Al}$

(d) By ${\text{NaClO}}_{3}$

${\text{NaClO"_3 + "2Al" → "NaCl" + "Al"_2"O}}_{3}$

$\text{Mass of Al" = 70 color(red)(cancel(color(black)("g NaClO"_3))) × (1 color(red)(cancel(color(black)("mol NaClO"_3))))/("106,44" color(red)(cancel(color(black)("g NaClO"_3)))) × (2 color(red)(cancel(color(black)("mol Al"))))/(1 color(red)(cancel(color(black)("mol NaClO"_3)))) × "26,98 g Al"/(1 color(red)(cancel(color(black)("mol Al")))) = "35,5 g Al}$

Total mass of $\text{Al}$ consumed

$\text{Mass of Al" = "(156,5 + 33,62 + 32,4 + 35,5) g Al" = "258 g Al}$