Question

A 0.853-g of 90.0% lithium chlorate mixture decomposes with heat to give 313 ml of oxygen gas collected over water at 20 c and 762 mmHg .what is the percentage of LiCLO3 in the mixture?

Answer 1

`\%LiClO_3=90.4\%`

Step-by-step explanation:

Hello!

In this case, since the undergoing chemical reaction is:

`LiClO_4\rightarrow LiCl+(3)/(2) O_2`

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:

`p_(O_2)=762mmHg-17.5mmHg=744.5mmHg*(1atm)/(760mmHg)=0.980atm`

Next, by using the ideal gas equation we compute the yielded moles of oxygen considering the collected 313 mL (0.313 L):

`n_(O_2)=(PV)/(RT)=(0.980atm*0.313L)/(0.082(atm*L)/(mol*K)*293.15K)\\\\ n_(O_2)=0.0128molO_2`

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:

`m_(LiClO_3)=0.0128molO_2*(1molLiClO_3)/(3/2molO_2)*(90.39gLiClO_3)/(1molLiClO_3) \\\\m_(LiClO_3)=0.771gLiClO_3`

Now, the percentage is computed as shown below:

`\%LiClO_3=(0.771g)/(0.853g) *100\%\\\\\%LiClO_3=90.4\%`

Best regards!