Answered

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

Sagot :

Answer: [tex]S[/tex] is the limiting reagent

Explanation:

To calculate the moles :

[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]    

[tex]\text{Moles of} S=\frac{10.0g}{32g/mol}=0.3125moles[/tex]

[tex]\text{Moles of} Ag=0.700moles[/tex]

[tex]2Ag+S\rightarrow Ag_2S[/tex]  

According to stoichiometry :

1 mole of [tex]S[/tex] require = 2 moles of [tex]Ag[/tex]  

Thus 0.3125 moles of [tex]S[/tex] will require=[tex]\frac{2}{1}\times 0.3125=0.6250moles[/tex]  of [tex]Ag[/tex]

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

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