Question

Grams and moles help please, have a test tomorrow on these questions.

1) How many grams of precipitate will be formed when 45.5 mL of 0.300 M Na₃PO₄ reacts with 50.0 mL of 0.200 M Cr(NO₃)₃ in the following chemical reaction? Na₃PO₄ (aq) + Cr(NO₃)₃ (aq) → CrPO₄ (s) + 3 NaNO₃ (aq)

2) How many moles of precipitate are formed when 20.0 mL of 0.250 M Ca(NO₃)₂ is mixed with excess Li₃PO₄ in the following chemical reaction? 3 Ca(NO₃)₂ (aq) + 2 Li₃PO₄ (aq) → Ca₃(PO₄)₂ (s) + 6 LiNO₃ (aq)

Answer 1

1) Moles of Na₃PO₄ = molarity * volume = 0.30 * 45.5 = 13.65 milli moles

Moles of Cr(NO₃)₃ = 0.20 * 50 = 10 milli moles

From the equation, we see that moles of Na₃PO₄ needed = moles of Cr(NO₃)₃

However since there is fewer milli moles of Cr(NO₃)₃ , it is the limiting reagent.

Moles of Precipitate (CrPO₄) = Moles of Cr(NO₃)₃ reacted = 10 milli moles

Mass of precipitate = moles * molar mass = 10 * 10^-3 * 147 = 1.47 g

2) Moles of Ca(NO₃)₂ = molarity * Volume = 0.25 * 20 = 5 milli moles

From the equation, moles of Precipitate formed (Ca₃(PO₄)₂) = 1 / 3 * moles of Ca(NO₃)₂ reacted = 1.667 milli moles

Mass of Ca₃(PO₄)₂ = moles * molar mass = 1.667 * 10^-3 * 310.2 = 0.517 g