89.2k views
2 votes
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

User Jeff Mc
by
7.9k points

2 Answers

1 vote

Answer:

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)

User MutableVoid
by
8.4k points
5 votes

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.

User Dirk Mahler
by
7.7k points