Respuesta :
0.375 g of [tex]CaF_2[/tex] will be produced from the given masses of both reactants.
Explanation:
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}[/tex] .....(1)
Molarity of [tex]KF[/tex] solution = 0.612 M
Volume of [tex]KF[/tex] solution = 15.7 mL
Putting values in equation 1, we get:
[tex]0.612M=\frac{\text{Moles of KF}\times 1000}{15.7ml}\\\\\text{Moles of KF}=\frac{0.612mol/L\times 15.7}{1000}=9.61\times 10^{-3}mol[/tex]
The balanced chemical reaction is:
[tex]2KF+Ca(HCO_3)_2\rightarrow CaF_2+2KHCO_3[/tex]
As [tex]KF[/tex] is the limiting reagent as it limits the formation of product and [tex]Ca(HCO_3)_2[/tex] is the excess reagent.
According to stoichiometry :
2 moles of [tex]KF[/tex] give = 1 mole of [tex]CaF_2[/tex]
Thus [tex]9.61\times 10^{-3}mol[/tex] moles of [tex]KF[/tex] will require=[tex]\frac{1}{2}\times 9.61\times 10^{-3}mol=4.80\times 10^{-3}mol[/tex] of [tex]CaF_2[/tex]
Mass of [tex]CaF_2=moles\times {\text {Molar mass}}=4.80\times 10^{-3}moles\times 78.07g/mol=0.375g[/tex]
Learn More about stoichiometry
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