Question

Calculate the volume and the number of moles of oxygen libertated at stp when 5.2gm of sodium peroxide reacts with water

Answer 1

moles=5.2/40

so the volume =5.2/40×22.4

Answer 2

The volume and number of moles of oxygen gas liberated are 0.75 L and 0.0335 moles

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of sodium peroxide = 5.2 g

Molar mass of sodium peroxide = 78 g/mol

Putting values in above equation, we get:

$\text{Moles of sodium peroxide}=\frac{5.2g}{78g/mol}=0.067mol$

The chemical equation for the reaction of sodium peroxide with water follows:

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

By Stoichiometry of the reaction:

2 moles of sodium peroxide produces 1 mole of oxygen gas

So, 0.067 moles of sodium peroxide will produce = $\frac{1}{2}\times 0.067=0.0335mol$ of oxygen gas

At STP conditions:

1 mole of a gas occupies 22.4 L of volume

So, 0.0335 moles of oxygen gas will occupy = $\frac{22.4}{1}\times 0.0335=0.75L$ of volume

Learn more about STP and stoichiometry:

https://brainly.com/question/2293072

https://brainly.in/question/13134014

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