Question

What volume of 0.250 M AgNO₃(aq) is required to react with 35.0 mL of 0.210 M CaBr₂(aq) in the reaction below:
  2 AgNO₃(aq) + CaBr₂ (aq) → Ca(NO₃)₂(aq) + 2AgBr(s)

Answer 1

  Solution :

Given data

0.250 M AgNO3

Volume of CaBr2 = 35.0 ml * 0.0350 L

Concentration of the CaBr2 = 0.210 M

Volume of AgNO3 = ?

2 AgNO₃(aq) + CaBr₂ (aq) ---- > Ca(NO₃)₂(aq) + 2AgBr(s)

Lets first calculate the moles of the CaBr2 using its molarity and volume

Moles = molarity * volume in liter

Moles of CaBr2 = 0.210 mol per L * 0.0350 L

                             = 0.00735 mol CaBr2

Now using the mole ratio of the CaBr2 and AgNO3 lets calculate the moles of AgNO3

Mole ratio of the CaBr2 to AgNO3 is 1 : 2

(0.00735 mol CaBr2 * 2 mol AgNO3) / 1 mol CaBr2 = 0.0147 mol AgNO3

So we need 0.0147 mol AgNO3

Now lets calculate the volume of AgNO3 using the molarity and moles

Volume in liter = moles / molarity

Volume in liter of AgNO3 = 0.0147 mol / 0.250 mol per L

                                              = 0.0588 L

Now lets convert this volume from liter to ml

0.0588 L * 1000 ml / 1 L = 58.8 ml

Therefore volume of the 0.250 M AgNO3 needed is 58.8 ml to react with 35.0 ml of 0.210 M CaBr2