A sloution of calcium nitrate, C(NO3), reacts with a solution of ammonium fluoride, NH4F, to form solid calcium fluoride, CaF2, and ammonium nitrate, NH4NO3(ag). When 45.00 mL of 4.8724 x 10 -1M Ca(NO3)2 solution was added to 60.00 mL of 9.9981 x 10 -1M NH4F solution, 1.524g of CaF2 was isolated.

(1) Write the equation for the reaction, showing the staes of substances.

(2) Calculate the number of moles of Ca(NO3)2 orginally present.

(3) calculate the number of moles of NH4F present

(4) Identify the limiting reagent.

(5) Calculate the theoretical yeild, in grams, of CaF2.

(6) Calculate the percent yield of CaF2.

You will have to calculate the numbers of oxidations in order to prove this to be a REDOX-reaction... The rest is simple algebra.

Yes it may be simple to you but it is not easy for me. Im not even sure where to begin! Please provide more in depth help!

(1) The balanced chemical equation for the reaction is:

Ca(NO3)2 + 2NH4F -> CaF2 + 2NH4NO3

(2) To calculate the number of moles of Ca(NO3)2 originally present, we need to use the concentration and volume of the solution. The formula to calculate the number of moles is:

moles = concentration (M) x volume (L)

Given:
Concentration of Ca(NO3)2 = 4.8724 x 10^(-1) M
Volume of Ca(NO3)2 solution = 45.00 mL = 0.04500 L

moles of Ca(NO3)2 = (4.8724 x 10^(-1) M) x (0.04500 L) = 2.1926 x 10^(-2) moles

(3) Similarly, to calculate the number of moles of NH4F present, we use the concentration and volume of the solution.

Given:
Concentration of NH4F = 9.9981 x 10^(-1) M
Volume of NH4F solution = 60.00 mL = 0.06000 L

moles of NH4F = (9.9981 x 10^(-1) M) x (0.06000 L) = 5.9989 x 10^(-2) moles

(4) To determine the limiting reagent, we compare the moles of Ca(NO3)2 and NH4F. The reactant that is completely consumed in the reaction is the limiting reagent.

From (2), moles of Ca(NO3)2 = 2.1926 x 10^(-2) moles
From (3), moles of NH4F = 5.9989 x 10^(-2) moles

Since 1 mole of Ca(NO3)2 reacts with 2 moles of NH4F, we need to compare the moles of NH4F with half of the moles of Ca(NO3)2.

Half moles of Ca(NO3)2 = (2.1926 x 10^(-2) moles) / 2 = 1.0963 x 10^(-2) moles

Since moles of NH4F > half moles of Ca(NO3)2, NH4F is the limiting reagent.

(5) To calculate the theoretical yield of CaF2, we use the stoichiometry of the balanced equation. As per the equation:

1 mole of Ca(NO3)2 produces 1 mole of CaF2

Therefore, moles of CaF2 = moles of Ca(NO3)2 = 1.0963 x 10^(-2) moles

The molar mass of CaF2 = 40.08 g/mol + 2 x 19.00 g/mol = 78.08 g/mol

Theoretical yield = moles of CaF2 x molar mass of CaF2
Theoretical yield = (1.0963 x 10^(-2) moles) x (78.08 g/mol) = 0.8564 g

(6) Finally, to calculate the percent yield of CaF2, we need the actual yield of CaF2, which is given as 1.524 g.

Percent yield = (Actual yield / Theoretical yield) x 100
Percent yield = (1.524 g / 0.8564 g) x 100 = 177.9% (rounded to three significant figures)

Note: The calculated percent yield is greater than 100% because there might have been experimental errors or impurities that resulted in a higher yield than expected.