Question

1.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

Answer: 588ml or 0.588Liters

2 AgNO₃(aq)   + CaBr₂ (aq)    ® Ca(NO₃)₂(aq) + 2AgBr(s)  

from the reaction given, it is clear that two moles of silver nitrate react with one mole of calcium bromide.

If we calculate the number of moles of silver nitrate and calcium bromide from their respective solutions given, we can find out the answer.

Number of moles of CaBr2:

concentration given = 0.210M or 0.210mol/L

that means there are 0.210 moles present in 1Liter of the solution.

So the number of moles of CaBr2 present in given 35 ml or 0.035 liter solution = ( 0.035 / 1) x 0.210

= 0.00735 oles.

Number of moles of AgNO3 reacting with CaBr2 = 2 x .00735 = 0.0147 moles.

Concentration given for the AgNO3 solution = 0.250M or 0.250 mol/Liter

That means there are 0.250 mole of AgNO3 present in one Liter solution.

So the volume of the same solution which has 0.0147 moles = (0.0147 / 0.250 ) x 1 = 0.0588 liters

or 588 ml of 0.250M AgNO3 solution is requires to react with 35.0 mL of 0.210 M CaBr₂(aq)