Asked by X
When 10.8 g of Silver was reacted with Sulfur, 12.4 grams of product was produced (there was only one product). What is the empirical formula of the product?
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Answered by
DrBob222
2Ag + S ==> Ag2S
If you started with 10.8 g Ag and ended up with 12.4g Ag2S, that means you must have added 1.6 g S. Convert these masses to mols by mols = grams/atomic mass
mols Ag = 10.8/estimated 108 = estimated 0.1
molsl S = 1.6/estimated 32 = 0.05
Now fine the ratio of these atoms to one another with the smallest whole number being 1.00 The easy way to do that is to divide the smaller number by itself and divide the other number by the same small number.
S = 0.05/0.05 = 1.00
Ag = 0.1/0.05 = 2.00
So the formula is .....
If you started with 10.8 g Ag and ended up with 12.4g Ag2S, that means you must have added 1.6 g S. Convert these masses to mols by mols = grams/atomic mass
mols Ag = 10.8/estimated 108 = estimated 0.1
molsl S = 1.6/estimated 32 = 0.05
Now fine the ratio of these atoms to one another with the smallest whole number being 1.00 The easy way to do that is to divide the smaller number by itself and divide the other number by the same small number.
S = 0.05/0.05 = 1.00
Ag = 0.1/0.05 = 2.00
So the formula is .....
Answered by
Anonymous
ag2s
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