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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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