Question

How many grams of the salt CaF2 (g) are formed when 15.7 mL of 0.612 M KF reacts with an excess of aqueous calcium bicarbonate (Ca(HCO3)2) via a metathesis reaction?

Answer 1

`m_(CaF_2)0.375gCaF_2`

Step-by-step explanation:

Hello,

In this case, for the studied reaction:

`2KF+Ca(HCO_3)_2\rightarrow CaF_2+2KHCO_3`

Thus, the first step is to compute the reacting moles of potassium fluoride by using its volume and molarity:

`n_(KF)=0.0157L*0.612(mol)/(L) =9.61x10^(-3)molKF`

Then, we apply the 2:1 molar ratio between potassium fluoride and calcium fluoride to compute the produced moles of calcium fluoride:

`n_(CaF_2)=9.61x10^(-3)molKF*(1molCaF_2)/(2molKF) =4.80x10^(-3)molCaF_2`

Finally, by using the molar mass of calcium fluoride (78.07 g/mol) we can compute its produced grams:

`m_(CaF_2)=4.80x10^(-3)molCaF_2*(78.07gCaF_2)/(1molCaF_2) \\\\m_(CaF_2)0.375gCaF_2`

Best regards.