What mass (g) of CaF₂is formed when 47.8 mL of 0.334 M NaF is treated with an excess of aqueous calcium nitrate? A) 2.49 B) 0.472 C) 1.25 D) 0.623 E) 0.943

Question

Quizplus · Accepted Answer

First, write a balanced chemical equation:

2 NaF + Ca(NO3)2 → CaF2 + 2 NaNO3

This reaction shows that 2 moles of NaF react with 1 mole of Ca(NO3)2 to produce 1 mole of CaF2.

Next, use the given concentration and volume of NaF to find the number of moles of NaF:

0.334 M NaF x 0.0478 L = 0.01594 mol NaF

Since NaF and Ca(NO3)2 react in a 2:1 mole ratio, we know that 0.00797 mol Ca(NO3)2 was used in the reaction (assuming excess Ca(NO3)2 is present).

Finally, use the molar mass of CaF2 to convert moles to grams:

0.00797 mol CaF2 x 78.07 g/mol CaF2 = 0.623 g CaF2

Therefore, the answer is D) 0.623.

What Mass (G) of CaF2 is Formed When 47

What Mass (G) of CaF₂is Formed When 47