Question

The decomposition of potassium chlorate yields oxygen gas. If the yield is 85%, how many grams of KClO3 are needed to produce 15.0 L of O2 ?
2KClO3(s) -- > 2KCl(s) + 3 O2(g)

Answer 1

Answer:

$m_{KClO_3}=64.4gKClO_3$

Explanation:

Hello,

At first, the undergoing chemical reaction is:

$2KClO_3(s)-->2KCl(s)+3O_2(s)$

As 15.0 L of oxygen were produced, one computes this amount in moles considering STP conditions (0°C and 1 atm) for the ideal gas equation as shown below:

$n_{O_2}=\frac{PV}{RT} =\frac{1atm*15.0L}{0.082 \frac{atm*L}{mol*K} *273.15K}=0.670molO_2$

Now, since those moles are the actual obtained moles, one computes the theoretical moles of oxygen which are more precisely associated with the needed grams of potassium chlorate as follows:

$n_{O_2}^{Theoretical}=\frac{0.670molO_2}{0.85} =0.788molO_2$

Finally, we apply the stoichiometry to determine the moles of potassium chlorate as shown below:

$m_{KClO_3}=0.788molO_2*\frac{2molKClO_3}{3molO_2} *\frac{122.55gKClO_3}{1molKClO_3} \\m_{KClO_3}=64.4gKClO_3$

Best regards.

Answer 2

 2KClO3 = 2KCl + 3O2. 

10 L are 10/22.4 moles as 1 mole of gas at STP occupies 22.4 L so 0.446 moles but you only get 95% yield so you need to make more than 0.446 moles -- you need 0.446/0.95 = 0.470 moles. 

Two moles KClO3 give 3 moles O2 so you need 0.470*(2/3) moles KClO3 = 0.313 moles KClO3 and that has molar mass = 39 + 35.5 + 3*16 = 122.5 g so in g 0.313*122.5 = 38.4 g KClO3