Respuesta :
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:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of lithium chlorate = 1.115 g
Molar mass of lithium chlorate = 90.39 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of lithium chlorate}=\frac{1.115g}{90.39g/mol}=0.0123mol[/tex]
At STP:
1 mole of a gas occupies 22.4 L of volume.
So, 0.0123 moles of lithium chlorate will occupy [tex]22.4\times 0.0123=0.27552L[/tex] of volume.
The chemical reaction for the decomposition of lithium chlorate follows the equation:
[tex]2LiClO_3\rightarrow 2LiCl+3O_2[/tex]
By Stoichiometry of the reaction:
[tex](2\times 22.4L)[/tex] of lithium chlorate produces [tex](3\times 22.4L)[/tex] of oxygen gas.
So, 0.27552 L of lithium chlorate will produce = [tex]\frac{(3\times 22.4)L}{(2\times 22.4)L}\times 0.27552=0.41328L[/tex] 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.
