Ask your own question, for FREE!
Chemistry 15 Online
OpenStudy (lovelyharmonics):

If 59.3 grams of potassium chlorate decomposes, how many grams of oxygen gas can be produced? 2 KClO3 → 2 KCl + 3 O2

OpenStudy (lovelyharmonics):

@ganeshie8

OpenStudy (lovelyharmonics):

My options are 15.5 g O2 23.2 g O2 31.0 g O2 46.4 g O2

ganeshie8 (ganeshie8):

start by finding the molar mass of KCLO3 : K : 39.1 Cl : 35.45 O3 : 16*3 = 48 ----------------- KCLO3 : 39.1 + 35.45 + 48 = 122.55 g/mol

ganeshie8 (ganeshie8):

tell me, how many mols are present in 59.3 grams of KCLO3 ?

OpenStudy (lovelyharmonics):

2.1

ganeshie8 (ganeshie8):

Nope. 122.55 g has 1 mol how can 59.3g be more than 1 mol ?

OpenStudy (lovelyharmonics):

.-. 122.55/59.3..... 59.3+59.3=118.6

OpenStudy (lovelyharmonics):

118 fits into 122

ganeshie8 (ganeshie8):

\[\large \text{59.3 g} = \text{59.3 g} \times \dfrac{\text{1 mol }}{\text{122.55 g}} = 0.484 ~\text{mol}\]

OpenStudy (lovelyharmonics):

.-. Wtf.... Where did that just come from

ganeshie8 (ganeshie8):

`KCLO3 : 39.1 + 35.45 + 48 = 122.55 g/mol` 1 mol = 122.55 g

ganeshie8 (ganeshie8):

? mol = 59.3 g

ganeshie8 (ganeshie8):

thats the question.

ganeshie8 (ganeshie8):

lets try this : 1 mol = 122.55 g so, 1g = 1/122.55 mol right ?

OpenStudy (lovelyharmonics):

Yus

ganeshie8 (ganeshie8):

finally a yes xD 1g = 1/122.55 mol multiply both sides by 59.3, what do u get ?

OpenStudy (lovelyharmonics):

0.484

ganeshie8 (ganeshie8):

1g = 1/122.55 mol 59.3 g = 59.3/122.55 mol = 0.484 mol

ganeshie8 (ganeshie8):

that means you have `0.484 mol of KCLO3 ` in the reaction, lets see how much O2 this produces

OpenStudy (lovelyharmonics):

And how do I do that .-.

ganeshie8 (ganeshie8):

2 KClO3 → 2 KCl + 3 O2 KClO3 → KCl + 3/2 O2

ganeshie8 (ganeshie8):

multiply through by 0.484

OpenStudy (lovelyharmonics):

.-. Multiply what through

ganeshie8 (ganeshie8):

0.484KClO3 → 0.484 KCl + `0.484*3/2` O2

ganeshie8 (ganeshie8):

So it produces 0.484*3/2 mol of O2

ganeshie8 (ganeshie8):

convert mols to grams, and you're done!

ganeshie8 (ganeshie8):

0.484*3/2 mol of O2 = How many grams ?

OpenStudy (lovelyharmonics):

0.726 c:

ganeshie8 (ganeshie8):

yes 0.726 mol of O2 = How many grams ?

ganeshie8 (ganeshie8):

whats the molar mass of O2 ?

OpenStudy (lovelyharmonics):

.-. I have no idea

ganeshie8 (ganeshie8):

google

ganeshie8 (ganeshie8):

wolfram http://www.wolframalpha.com/input/?i=molar+mass+O2

OpenStudy (lovelyharmonics):

N c: just use your stuff from up top. 16*2=32

OpenStudy (lovelyharmonics):

*no

ganeshie8 (ganeshie8):

that works ! 1 mol of O2 = 32 grams

ganeshie8 (ganeshie8):

0.726 mol of O2 = How many grams ?

OpenStudy (lovelyharmonics):

1/32=0.726/? 232.32/1

OpenStudy (lovelyharmonics):

I'm guessing mines wrong and the correct answer is 23.2

ganeshie8 (ganeshie8):

tell me this : if 1 coconut weighs 1 kg then half coconut weighs how many kg ?

OpenStudy (lovelyharmonics):

That's the great thing about multiple choice right there XD

ganeshie8 (ganeshie8):

23.2 is correct !

OpenStudy (lovelyharmonics):

.5kg....

ganeshie8 (ganeshie8):

how did u get .5kg ?

ganeshie8 (ganeshie8):

you multiplied the original weight by 1/2 right ?

OpenStudy (lovelyharmonics):

By halfing it .-.

OpenStudy (lovelyharmonics):

Well I don't really need to go through the whole process of 1*1/2=.5

OpenStudy (lovelyharmonics):

But yeah I guess what your sayin

ganeshie8 (ganeshie8):

then tell me this, how much 0.235 of coconut weighs ?

ganeshie8 (ganeshie8):

how much \(\large \pi \) of coconut(s) weigh ?

ganeshie8 (ganeshie8):

how much 10 coconuts weigh ?

ganeshie8 (ganeshie8):

this is essentially a multiplication problem, not a division problem. okay ?

OpenStudy (lovelyharmonics):

Yeah it's cross multiplication that's what I did

ganeshie8 (ganeshie8):

okay nice :) anything is fine as long as it makes sense...

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!