Answer:
[tex]m_{CaF_2}0.375gCaF_2[/tex]
Explanation:
Hello,
In this case, for the studied reaction:
[tex]2KF+Ca(HCO_3)_2\rightarrow CaF_2+2KHCO_3[/tex]
Thus, the first step is to compute the reacting moles of potassium fluoride by using its volume and molarity:
[tex]n_{KF}=0.0157L*0.612\frac{mol}{L} =9.61x10^{-3}molKF[/tex]
Then, we apply the 2:1 molar ratio between potassium fluoride and calcium fluoride to compute the produced moles of calcium fluoride:
[tex]n_{CaF_2}=9.61x10^{-3}molKF*\frac{1molCaF_2}{2molKF} =4.80x10^{-3}molCaF_2[/tex]
Finally, by using the molar mass of calcium fluoride (78.07 g/mol) we can compute its produced grams:
[tex]m_{CaF_2}=4.80x10^{-3}molCaF_2*\frac{78.07gCaF_2}{1molCaF_2} \\\\m_{CaF_2}0.375gCaF_2[/tex]
Best regards.
