<h2>0.375 g of
![CaF_2](https://tex.z-dn.net/?f=CaF_2)
will be produced from the given masses of both reactants.</h2>
Explanation:
To calculate the number of moles for given molarity, we use the equation:
.....(1)
Molarity of
solution = 0.612 M
Volume of
solution = 15.7 mL
Putting values in equation 1, we get:
![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](https://tex.z-dn.net/?f=0.612M%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20KF%7D%5Ctimes%201000%7D%7B15.7ml%7D%5C%5C%5C%5C%5Ctext%7BMoles%20of%20KF%7D%3D%5Cfrac%7B0.612mol%2FL%5Ctimes%2015.7%7D%7B1000%7D%3D9.61%5Ctimes%2010%5E%7B-3%7Dmol)
The balanced chemical reaction is:
![2KF+Ca(HCO_3)_2\rightarrow CaF_2+2KHCO_3](https://tex.z-dn.net/?f=2KF%2BCa%28HCO_3%29_2%5Crightarrow%20CaF_2%2B2KHCO_3)
As
is the limiting reagent as it limits the formation of product and
is the excess reagent.
According to stoichiometry :
2 moles of
give = 1 mole of
Thus
moles of
will require=
of
Mass of
Learn More about stoichiometry
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