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If 0.700 moles Ag is reacted with 10.0 g S, is sulfur or aluminum the limiting reactant?

User Mbbce
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1 Answer

2 votes

Answer:
S is the limiting reagent

Step-by-step explanation:

To calculate the moles :


\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}


\text{Moles of} S=(10.0g)/(32g/mol)=0.3125moles


\text{Moles of} Ag=0.700moles


2Ag+S\rightarrow Ag_2S

According to stoichiometry :

1 mole of
S require = 2 moles of
Ag

Thus 0.3125 moles of
S will require=
(2)/(1)* 0.3125=0.6250moles of
Ag

Thus
S is the limiting reagent as it limits the formation of product and
Ag is the excess reagent.

User Smassey
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