Answer:
[tex]\%LiClO_3=90.4\%[/tex]
Explanation:
Hello!
In this case, since the undergoing chemical reaction is:
[tex]LiClO_4\rightarrow LiCl+\frac{3}{2} O_2[/tex]
It is widely known that when a gas given off from a reaction is collected over water, we can compute its pressure by minusing the total pressure 762 mmHg and the vapor pressure of water at the experiment's temperature (20 °C) in this case 17.5 mmHg as shown below:
[tex]p_{O_2}=762mmHg-17.5mmHg=744.5mmHg*\frac{1atm}{760mmHg}=0.980atm[/tex]
Next, by using the ideal gas equation we compute the yielded moles of oxygen considering the collected 313 mL (0.313 L):
[tex]n_{O_2}=\frac{PV}{RT}=\frac{0.980atm*0.313L}{0.082\frac{atm*L}{mol*K}*293.15K}\\\\ n_{O_2}=0.0128molO_2[/tex]
Now, via the 3/2:1 mole ratio between oxygen and lithium chlorate (molar mass = 90.39 g/mol), we compute the original mass of decomposed lithium chlorate as follows:
[tex]m_{LiClO_3}=0.0128molO_2*\frac{1molLiClO_3}{3/2molO_2}*\frac{90.39gLiClO_3}{1molLiClO_3} \\\\m_{LiClO_3}=0.771gLiClO_3[/tex]
Now, the percentage is computed as shown below:
[tex]\%LiClO_3=\frac{0.771g}{0.853g} *100\%\\\\\%LiClO_3=90.4\%[/tex]
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