# 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. $13 \space moles \space O_2$

#### Work Step by Step

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. $$6.4 \space moles \space KClO_4 \times \frac{ 2 \space moles \ O_2 }{ 1 \space mole \space KClO_4 } = 13 \space moles \space O_2$$

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