A solution of calcium nitrate, Ca(NO3)2, reacts with a solution of ammonium fluoride, NH4F, to form solid calcium fluoride, CaF2, and ammonium nitrate, NH4NO3(aq). When 45.00 mL of 4.8724 x 10-1 M Ca(NO3)2 solution was added to 60.00 mL of 9.9981 x 10-1 M NH4F solution, 1.524 g of CaF2 was isolated.
Calculate the number of moles of Ca(NO3)2 originally present
See your post below. Same procedure.
0.5999
To calculate the number of moles of Ca(NO3)2 originally present, we need to use the given information about the volume and concentration of the Ca(NO3)2 solution. Here's how you can do it:
1. Convert the given volume of the Ca(NO3)2 solution to liters.
- 45.00 mL ÷ 1000 = 0.045 L
2. Calculate the number of moles using the formula:
moles = concentration × volume
- moles = 4.8724 x 10^-1 M × 0.045 L
3. Calculate the number of moles of Ca(NO3)2 originally present.
- moles = 4.8724 x 10^-1 M × 0.045 L = 2.19258 x 10^-2 moles
Therefore, the number of moles of Ca(NO3)2 originally present is approximately 2.19258 x 10^-2 moles.