What mass (g)of CaF₂ is formed when 30.4 mL of 0.438 M NaF is treated with an excess of aqueous calcium nitrate? A)0.520 B)1.04 C)1040 D)0.171 E)5420

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The balanced chemical equation is:2NaF + Ca(NO3)2 → CaF2 + 2NaNO3Moles of NaF = 0.438 M * 0.0304 L = 0.0133 molMoles of CaF2 = 0.0133 mol / 2 = 0.00665 molMass of CaF2 = 0.00665 mol * 78.07 g/mol = 0.520 g

What Mass (G)of CaF2 Is Formed When 30

What Mass (G)of CaF₂ Is Formed When 30