Question

Reaction amounts

Substance- Amount (reacted or formed)
KClO3-- 10.0 g
KCl- 7.5 g
Al- 9.0 g


Reaction 1 ___KClO3(s) → ___KCl(s) + ___O2(g)


Reaction 2 ___Al(s) + ___O2(g) → ___Al2O3(s)


The decomposition of potassium chlorate in reaction 1 provides the necessary oxygen for reaction 2 to take place.
Determine the coefficients that will balance reactions 1 & 2. Also, find the mass of Al2O3 that will form given the amounts in the table.
A) Reaction 1: 2, 2, and 3
Reaction 2: 2, 3, and 2
7.5 grams Al2O3
B) Reaction 1: 1, 1, and 2
Reaction 2: 2, 3, and 2
9.5 grams Al2O3
C) Reaction 1: 2, 2, and 3
Reaction 2: 4, 3, and 2
11.5 grams Al2O3
D) Reaction 1: 2, 2, and 3
Reaction 2: 2, 2, and 3
16.5 grams Al2O3

Answer 1

C) Reaction 1: 2, 2, and 3

Reaction 2: 4, 3, and 2

11.5 grams Al2O3

Step-by-step explanation:

Reaction 1. __2_KClO3(s) → __2_KCl(s) + __3_O2(g)

Reaction 2. __4_Al(s) + __3_O2(g) → __2_Al2O3(s)

Answer 2

c)

Reaction 1: 2, 2, and 3

Reaction 2: 4, 3, and 2

11.5 grams Al2O3

According to the law of conservation of matter and/or mass, 2.5 grams of O2 is formed in the first reaction. Using this amount with the given aluminum amount, 11.5 grams of grams Al2O3 will be formed in reaction 2.