Chapter 3 - Chemical Equations and Reaction of Stoichiometry - Exercises - Calculations Based on Chemical Equations - Page 107: 18

a. $ 9.6 \space moles \space O_2 $ b. $ 3.2 \space moles \space O_2 $ c. $ 3.2 \space moles \space O_2 $ d. $ 3.2 \space moles \space O_2 $ e. $ 13 \space moles \space O_2 $

a. $$ 6.4 \space moles \space KClO_3 \times \frac{ 3 \space moles \ O_2 }{ 2 \space moles \space KClO_3 } = 9.6 \space moles \space O_2 $$ b. $$ 6.4 \space moles \space H_2O_2 \times \frac{ 1 \space mole \ O_2 }{ 2 \space moles \space H_2O_2 } = 3.2 \space moles \space O_2 $$ c. $$ 6.4 \space moles \space HgO \times \frac{ 1 \space mole \ O_2 }{ 2 \space moles \space HgO } = 3.2 \space moles \space O_2 $$ d.$$ 6.4 \space moles \space NaNO_3 \times \frac{ 1 \space mole \ O_2 }{ 2 \space moles \space NaNO_3 } = 3.2 \space moles \space O_2 $$ e. $$ 6.4 \space moles \space KClO_4 \times \frac{ 2 \space moles \ O_2 }{ 1 \space mole \space KClO_4 } = 13 \space moles \space O_2 $$