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

Answer:

$\%LiClO_3=90.4\%$

Explanation:

Hello!

In this case, since the undergoing chemical reaction is:

$LiClO_4\rightarrow LiCl+\frac{3}{2} O_2$

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:

$p_{O_2}=762mmHg-17.5mmHg=744.5mmHg*\frac{1atm}{760mmHg}=0.980atm$

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

$n_{O_2}=\frac{PV}{RT}=\frac{0.980atm*0.313L}{0.082\frac{atm*L}{mol*K}*293.15K}\\\\ n_{O_2}=0.0128molO_2$

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:

$m_{LiClO_3}=0.0128molO_2*\frac{1molLiClO_3}{3/2molO_2}*\frac{90.39gLiClO_3}{1molLiClO_3} \\\\m_{LiClO_3}=0.771gLiClO_3$

Now, the percentage is computed as shown below:

$\%LiClO_3=\frac{0.771g}{0.853g} *100\%\\\\\%LiClO_3=90.4\%$

Best regards!