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.

Question

(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.

Answer 1

Ca(NO3)2(aq) + 2NH4F(aq) ==> CaF2(s) + 2NH4NO3(aq)

2. moles = M x L
3. moles = M x L
4. Using the coefficients in the balanced equation, convert mole of Ca(NO3)2 to moles CaF2. With the same procedure, convert moles NH4F to moles CaF2. The smaller number of moles will be the correct # moles for CaF2 and the reactant that produced that number will be the limiting reagent.
5. The theoretical yield will be obtained by using the smaller number of moles CaF2 and that times molar mass CaF2.
6. percent yield = (actual yield/theoretical yield)*100 = ??
theoretical yield you obtained in 5. Actual yield is given in the problem as 1.524 g CaF2.
Post your work if you get stuck.

Answer 2

gwsfd

Answer 3

(1) Ca(NO3)2(aq) + 2NH4F(aq) = CaF2(s) + 2NH4NO3 (aq)

(2) 2.1926 * 10^-2 mol Ca(NO3)2

(3) 5.9989 * 10^-2 mol NH4F

(4) Ca(NO3)2

(5) 1.71197 g CaF2

(6) 89.02%