Small quantities of oxygen can be prepared in the laboratory by heating potassium chlorate, KClO3(s). The equation for the reaction is

2KClO3⟶2KCl+3O2

Calculate how many grams of O2(g) can be produced from heating 66.7 g KClO3(s).

26.110 grams of O2 produced.

Explanation:

Calculate the amount of moles in KClO3 by dividing the amount of grams given by the atomic weight of the substance.

To get the atomic weight: K = 39.098, Cl = 35.45, O = 15.999, and there are 3 molecules of Oxygen, so multiply 15.999 by three.

39.098 + 35.45 + (15.999 * 3) = 122.548.

66.7g / 122.548 atomic mass = 0.544 moles.

The ratio of moles of KClO3 to moles of O2 is 2 to 3.

=

Cross multiply to get 1.632 = 2y. Y = 0.816, meaning 0.816 moles of O2 will be produced.

Convert this into grams by multiplying by the atomic weight of O2 (15.999 * 2 = 31.998).

0.816 * 31.998 = 26.110 grams of O2 produced.

i believe it is c.) 1575 j.