Question #5f09b

1 Answer
May 7, 2017

Answer:

WARNING! Long answer! The mass of #"Al"# to be used is 258 g.

Explanation:

The #"Al"# will react with the energizer and with every metal oxide in the mixture.

Mass of each reactant

#"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#

#"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"#

#"Mass of CrO"_3 = "60 g"#

#"Mass of KClO"_3 = "70 g"#

Mass of #"Al"# consumed by each reactant

(a) By #"Cr"_2"O"_3"#

#"Cr"_2"O"_3 + "2Al" → "2Cr" + "Al"_2"O"_3#

#"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 #"FeO"#

#"3FeO + 2Al" → "3Fe" + "Al"_2"O"_3#

#"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 #"CrO"_3#

#"CrO"_3 + "2Al" → "Cr" + "Al"_2"O"_3#

#"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 #"NaClO"_3#

#"NaClO"_3 + "2Al" → "NaCl" + "Al"_2"O"_3#

#"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 #"Al"# consumed

#"Mass of Al" = "(156,5 + 33,62 + 32,4 + 35,5) g Al" = "258 g Al"#