The required moles of AgBr precipitate produced by given moles of silver nitrate is 0.0123.
Molarity of any solution is define as the moles of solute present in per liter of the solution and it will be represented as:
M = n/V
Given that, molarity of AgNO₃ = 0.250M
Volume of AgNO₃ = 49.5mL = 0.0495L
Moles of AgNO₃ = (0.25)(0.0495) = 0.0123mol
Given chemical reaction is:
2AgNO₃(aq) + CaBr(aq) → 2AgBr(s) + Ca(NO₃)₂(aq)
As it is mention that CaBr is present in excess quantity and AgNO₃ is the limiting reagent so the formation of precipitate will depend on the AgNO₃.
From the stoichiometry of the reaction, it is clear that:
2 moles of AgNO₃ = produces 2 moles of AgBr
0.0123 moles of AgNO₃ = produces 2/2×0.0123=0.0123 moles of AgBr
Hence 0.0123 is the required moles of precipitate.
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