menu search
brightness_auto
more_vert

4) solve problems

P. Calculate the mass of potassium chlorate required to liberate 6.72 dm3 of oxygen at STP. Molar mass of KClO3 is 122.5 g mol-1

thumb_up_off_alt 0 like thumb_down_off_alt 0 dislike

1 Answer

more_vert
 
verified
Best answer

\(\begin{matrix} 2KClO_3 &\rightarrow 2KCl&+3O_2\\ 2moles & & 3 moles \end{matrix}\) 

2 moles of KClO3 = 2× 122.5=245g

3 moles of O2 =3× 22.4=67.2 dm3 

hence , 245 g of potacium chlorate will liberate 67.2 dm3 of oxygen

As , here  x gram of KClO3 liberate 6.72 dm of oxygen at stp 

\(\therefore x={245\times 6.72\over 67.2}=24.5 g\) 

required mass of potacium chlorate is 24.5 g

thumb_up_off_alt 0 like thumb_down_off_alt 0 dislike

Related questions

thumb_up_off_alt 1 like thumb_down_off_alt 0 dislike
1 answer
thumb_up_off_alt 0 like thumb_down_off_alt 0 dislike
1 answer
thumb_up_off_alt 0 like thumb_down_off_alt 0 dislike
1 answer
thumb_up_off_alt 0 like thumb_down_off_alt 0 dislike
1 answer

Welcome to Brainiak.in , where you can ask questions and receive answers from other members of the community ,

There is many categories for every students  CBSE , Maharashtra board , JEE , NEET  and we will add more categories

and make brainiak one of the most loved community of students

Advertisement

Get Your advertisement on Brainiak.in Buy Adspace

3.0k questions

2.5k answers

24 comments

341 users

...