How do I find the empirical formula of the product produced by heating 1 gram of Zinc Sulfur if i end up with 0,8375 grams of product ?

my formula is :

#X ZnS + ((2x+y)/2) O_2 rarr Zn_xO_y + XSO_2#

I determined the X by :

number of moles of Zn = number of moles ZnS (not sure why though)

gives :

#((1)/ (65,39)) = 0,0153#

I have no idea what I'm supposed to do with this information, or what I'm looking for now.

my formula is :

#X ZnS + ((2x+y)/2) O_2 rarr Zn_xO_y + XSO_2#

I determined the X by :

number of moles of Zn = number of moles ZnS (not sure why though)

gives :

#((1)/ (65,39)) = 0,0153#

I have no idea what I'm supposed to do with this information, or what I'm looking for now.

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devi f. Share
Mar 9, 2018

Answer:

By the way there is nothing called Zinc Sulpur. It is Zinc Sulphide There is no way for you to determine the product of the above reaction without knowing other properties of Zinc and Oxygen.

Explanation:

So you have Zinc Sulphide reacting with Oxygen to generate Zinc Oxide and Sulpur Dioxide . Assuming what you weigh is the Zinc Oxide only.

Can you have #Zn_xO_y# where x, y are anything other than 1?

Zinc has a valency of 2, Oygen has a valency of -2; Balanced, so you cannot have a compond other than #ZnO#

Your unbalanced equation would be:

#ZnS + O_2 > ZnO + SO_2#

Balance it using a simple count of number of atoms on each side:

LHS: 1 Zn , 1 S, 2 O
RHS: 1 Zn , 1 S, 3 O

Since the count of oxyen on both side is different, the equation is unbalanced, balance it to make values of the coefficeits same on both sides.

You get:

#2 ZnS + 3 O_2 = 2 ZnO + 2 SO_2#

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