Chapter 7 - Chemical Reactions and Quantities - Additional Questions and Problems - Page 284: 7.133

a. 3.00 moles of $H_2O_2$ are needed to produce 3.00 moles of $H_2O$ b. 77.6 g of $H_2O_2$ are required to produce 36.5 g of $O_2$ c. 6.46 g of $H_2O$ can be produced with 12.2 g of $H_2O_2$

a. $$ 3.00 \space moles \space H_2O \times \frac{ 2 \space moles \ H_2O_2 }{ 2 \space moles \space H_2O } = 3.00 \space moles \space H_2O_2 $$ b. $ O_2 $ : ( 16.00 $\times$ 2 )= 32.00 g/mol $$ \frac{1 \space mole \space O_2 }{ 32.00 \space g \space O_2 } \space and \space \frac{ 32.00 \space g \space O_2 }{1 \space mole \space O_2 }$$ $ H_2O_2 $ : ( 1.008 $\times$ 2 )+ ( 16.00 $\times$ 2 )= 34.02 g/mol $$ \frac{1 \space mole \space H_2O_2 }{ 34.02 \space g \space H_2O_2 } \space and \space \frac{ 34.02 \space g \space H_2O_2 }{1 \space mole \space H_2O_2 }$$ $$ 36.5 \space g \space O_2 \times \frac{1 \space mole \space O_2 }{ 32.00 \space g \space O_2 } \times \frac{ 2 \space moles \space H_2O_2 }{ 1 \space mole \space O_2 } \times \frac{ 34.02 \space g \space H_2O_2 }{1 \space mole \space H_2O_2 } = 77.6 \space g \space H_2O_2 $$ c. $ H_2O_2 $ : 34.02 g/mol $$ \frac{1 \space mole \space H_2O_2 }{ 34.02 \space g \space H_2O_2 } \space and \space \frac{ 34.02 \space g \space H_2O_2 }{1 \space mole \space H_2O_2 }$$ $ H_2O $ : ( 1.008 $\times$ 2 )+ ( 16.00 $\times$ 1 )= 18.02 g/mol $$ \frac{1 \space mole \space H_2O }{ 18.02 \space g \space H_2O } \space and \space \frac{ 18.02 \space g \space H_2O }{1 \space mole \space H_2O }$$ $$ 12.2 \space g \space H_2O_2 \times \frac{1 \space mole \space H_2O_2 }{ 34.02 \space g \space H_2O_2 } \times \frac{ 2 \space moles \space H_2O }{ 2 \space moles \space H_2O_2 } \times \frac{ 18.02 \space g \space H_2O }{1 \space mole \space H_2O } = 6.46 \space g \space H_2O $$