Answer to Question #107910 in Chemistry for Selena

Question

How many grams of KCIO3 are needed to make 1.5 x 10^24 particles of O2

2KCI + 3O2 = 2KCIO3

Answer 1

N(O₂) = 1.5*10²⁴ particles of O₂

Solution:

1)

N(O₂) = n(O₂) * N_a;

n(O₂) = N(O₂) / N_a;

n(O₂) = (1.5*10²⁴) / (6.02*10²³) = 2.4917 moles.

The equation of a chemical reaction:

2KCIO₃ = 2KCI + 3O₂

2)

According to the chemical equation: n(KCIO3)/2 = n(O2)/3.

n(KClO₃) = 2*n(O₂)/3;

n(KClO₃) = (2 * 2.4917 moles) / 3 = 1.66 moles.

3)

n(KClO₃) = m(KClO₃) / M(KClO₃);

M(KClO3) = Ar(K) + Ar(Cl) + 3*Ar(O) = 39 + 35.5 +3*16 = 122.5 (g/mol).

m(KClO₃) = n(KClO₃) * M(KClO₃) = (1.66 moles) * (122.5 g/mol) = 203.35 g.

m(KClO₃) = 203.35 g.

Answer: 203.35 grams of KClO₃.