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How many grams of #Ag# can be formed from 5.50 grams of #Ag_2O# in the equation: #2Ag_2O (s) -> 4Ag (s) + O_2 (g)#?

Answer 1

Naomi Aronson

Clearly #"moles of silver oxide:moles of dioxygen"# #=# #2:1#.

#"Moles of silver oxide"# #=# #(5.50*g)/(231.74*g*mol^-1)# #=# #??mol#

#"Grams of silver metal"# #=#

#2xx(5.50*g)/(231.74*g*mol^-1)xx107.87*g*mol^-1# #=# #??*g#.

Why did I put the #2# in the front of the expression?

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Answer 2

Wyatt Adkins

2.75 grams of Ag can be formed from 5.50 grams of Ag₂O.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How many grams of #Ag# can be formed from 5.50 grams of #Ag_2O# in the equation: #2Ag_2O (s) -&gt; 4Ag (s) + O_2 (g)#?

How many grams of #Ag# can be formed from 5.50 grams of #Ag_2O# in the equation: #2Ag_2O (s) -> 4Ag (s) + O_2 (g)#?