Question

Sodium peroxide (Na2O2) reacts with water to form sodium hydroxide and oxygen gas. Write a balanced equation for the reaction and determine how much oxygen in grams is formed by the complete reaction of 35.23 g of Na2O2.

Answer 1

Answer:
`2Na_2O_2+2H_2O\rightarrow 4NaOH+O_2`

7.229 grams of oxygen is formed by the complete reaction of 35.23 g of
`Na_2O_2`

Step-by-step explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

`2Na_2O_2+2H_2O\rightarrow 4NaOH+O_2`

To calculate the moles :

`\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}`

`\text{Moles of} Na_2O_2=(35.23g)/(77.98g/mol)=0.4518moles`

`2Na_2O_2+2H_2O\rightarrow 4NaOH+O_2`

According to stoichiometry :

2 moles of
`Na_2O_2` give = 1 mole of
`O_2`

Thus 0.4518 moles of
`Na_2O_2` give =
`(1)/(2)* 0.4518=0.2259moles` of
`O_2`

mass of
`O_2=moles* {\text {molar mass}}=0.2259mol* 32g/mol=7.229g`

7.229 grams of oxygen is formed by the complete reaction of 35.23 g of
`Na_2O_2`