The mass of CaF₂ formed when 47.8 mL of 0.334 M NaF is treated with an excess of aqueous calcium nitrate, Ca(NO₃)₂ is 0.624 g
We'll begin by calculating the number of mole of NaF in the solution. This can be obtained as follow:
Volume = 47.8 mL = 47.8 / 1000 = 0.0478 L
Molarity of NaF = 0.334 M
Mole of NaF =?
Mole = Molarity x Volume
Mole of NaF = 0.334 × 0.0478
Mole of NaF = 0.016 mole
- Next, we shall determine the number of mole of CaF produced from the reaction.
2NaF + Ca(NO₃)₂ —> CaF₂ + 2NaNO₃
NOTE: This is a double displacement reaction.
From the balanced equation above,
2 moles of NaF reacted to produce 1 mole of CaF₂.
Therefore,
0.016 mole of NaF will react to produce = 0.016 / 2 = 0.008 mole of CaF₂.
- Finally, we shall determine the mass of 0.008 mole of CaF₂.
Mole of CaF₂ = 0.008 mole
Molar mass of CaF₂ = 40 + (19×2) = 78 g/mol
Mass of CaF₂ =?
Mass = mole × molar mass
Mass of CaF₂ = 0.008 × 78
Mass of CaF₂ = 0.624 g
Therefore, the mass of CaF₂ formed when 47.8mL of 0.334 M NaF is treated with an excess of aqueous calcium nitrate, Ca(NO₃)₂ is 0.624 g
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