Answer :
Final answer:
The quantity in moles of precipitate formed when 51.7 mL of 0.250 M AgNO₃ is reacted with excess CaBr₂ in the chemical reaction 2 AgNO₃(aq) + CaBr₂(aq) → 2 AgBr(s) + Ca(NO₃)₂ is 0.012925 mol AgBr.
Explanation:
To calculate the quantity in moles of precipitate formed, we need to use the given information about the volume and concentration of silver nitrate (AgNO₃).
First, we convert the volume of AgNO₃ from milliliters to liters:
51.7 mL = 0.0517 L
Next, we use the concentration of AgNO₃ to calculate the number of moles:
0.0517 L * 0.250 mol/L = 0.012925 mol AgNO₃
According to the stoichiometry of the reaction, for every 2 moles of AgNO₃, 2 moles of AgBr are formed. Therefore, the quantity in moles of AgBr formed will be the same as the quantity in moles of AgNO₃:
0.012925 mol AgBr
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