Question

lithium chlorate is decomposed with heat to give lithium chloride and oxygen gas. If 1.115 g of lithium chlorate is decomposed, how many milliliters of oxygen gas is released at STP?

Answer 1

Answer: The volume of oxygen gas released at STP will be 413.28 mL.

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 lithium chlorate = 1.115 g

Molar mass of lithium chlorate = 90.39 g/mol

Putting values in above equation, we get:

$\text{Moles of lithium chlorate}=\frac{1.115g}{90.39g/mol}=0.0123mol$

At STP:

1 mole of a gas occupies 22.4 L of volume.

So, 0.0123 moles of lithium chlorate will occupy $22.4\times 0.0123=0.27552L$ of volume.

The chemical reaction for the decomposition of lithium chlorate follows the equation:

$2LiClO_3\rightarrow 2LiCl+3O_2$

By Stoichiometry of the reaction:

$(2\times 22.4L)$ of lithium chlorate produces $(3\times 22.4L)$ of oxygen gas.

So, 0.27552 L of lithium chlorate will produce = $\frac{(3\times 22.4)L}{(2\times 22.4)L}\times 0.27552=0.41328L$ of oxygen gas.

Converting the calculated volume into mililiters, we use the conversion factor:

1 L = 1000 mL

So, 0.41328 L = 0.41328 × 1000 = 413.28 mL

Hence, the volume of oxygen gas released at STP will be 413.28 mL.