The required moles of AgBr precipitate produced by given moles of silver nitrate is 0.0123.
<h3>How do we calculate moles from molarity?</h3>
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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