2 AgNO3(aq) + CaBr2(aq) → 2 AgBr(s) + Ca(NO3)2. What volume (mL) of a 0.125 M CaBr2 solution would be required to react completely with 50.0 mL of 0.115 M AgNO3?

Answer 1

2 AgNO₃(aq) + CaBr₂(aq) ==> 2 AgBr(s) + Ca(NO₃)₂ ... balanced equation

First we find the moles of AgNO₃ that are present:

50.0 ml x 1 L/1000 ml x 0.115 mol/L = 0.00575 moles AgNO₃ present

Next we find the number of moles of CaBr₂ are needed to react with this amount of AgNO₃:

0.00575 mol AgNO₃ x 1 mol CaBr₂ / 2 mol AgNO₃ = 0.002875 moles CaBr₂ needed

Finally, from the fact that the CaBr₂ is 0125 M = 0.125 mol/L, we can find the volume needed:

(x L)(0.125 mol/L) = 0.002875 moles

x = 0.023 L = 23.0 mls of CaB₂ are needed