How many milliliters of 0.200 M FeCl3 are needed to react with an excess of Na2S to produce 0.172 g of Fe2S3 if the percent yield for the reaction is 65.0%?3 Na2S(aq) + 2 FeCl3(aq) → Fe2S3(s) + 6 NaCl(aq)

What mass of precipitate (in g) is formed when 45.5 mL of 0.300 M Li₃PO₄ reacts with 56.0 mL of 0.200 M CrCl₃ in the following chemical reaction? Li₃PO₄(aq) + CrCr₃(aq) → CrPO₄(s) + 3 LiCl(aq)

What is the mass in grams of precipitate formed when 45.5 mL of 0.300 M Na₃PO₄ reacts with 59.0 mL of 0.200 M Cr(NO₃)₃ in the following chemical reaction? Na₃PO₄ (aq) + Cr(NO₃)₃ (aq) → CrPO₄ (s) + 3 NaNO₃ (aq)

You have 0.765 g of an unknown acid, H₂A, which reacts with NaOH according to the balanced equation H₂A(aq) + 2 NaOH(aq) → Na2 A(aq) + 2 H₂O(l) If 25.78 mL of 0.461 M NaOH is required to titrate the acid to the second equivalence point, what is the molar mass of the acid? Molar mass = g/mol

What mass of precipitate (in g) is formed when 20.5 mL of 0.800 M Co(NO₃)₂ reacts with 22.5 mL of 0.800 M NaOH in the following chemical reaction? Co(NO₃)₂(aq) + 2 NaOH(aq) → Co(OH)₂(s) + 2 NaNO₃(aq)

How many moles of precipitate will be formed when 75.0 mL of 2.00 M NaCl is reacted with 200.0 mL of 0.600 M Pb(NO₃)₂ in the following chemical reaction? 2 NaCl(aq) + Pb(NO₃)₂(aq) → PbCl₂(s) + 2 NaNO₃(aq)

What mass of precipitate (in g) is formed when 45.5 mL of 0.300 M K₃PO₄ reacts with 54.5 mL of 0.200 M CrCl₃ in the following chemical reaction? K₃PO₄(aq) + CrCl₃(aq) → CrPO₄(s) + 3 KCl₃(aq)

How many moles of precipitate are formed when 25.7 mL of 0.400 M Ca(NO₃)₂ is mixed with excess K₃PO₄ in the following chemical reaction? 3 Ca(NO₃)₂(aq) + 2 K₃PO₄(aq) → Ca₃(PO₄)₂(s) + 6 KNO₃(aq)

A 0.313-g sample of oxalic acid (H2C2O4), requires 26.21 mL of a particular concentration of NaOH(aq) to complete the following reaction. What is the molarity of the NaOH(aq)?H2C2O4(s) + 2NaOH(aq) → Na2C2O4(aq) + 2H2O(l)

Determine the mass (in g) of AgBr that is produced when 609 mL of a 5.76×10-2 M Na3Ag(S2O3)2 solution completely reacts with 674 mL of a 1.15×10-2 M NaBr solution according to the following balanced chemical equation.Na3Ag(S2O3)2(aq) + NaBr(aq) → AgBr(s) + 2Na2S2O3(aq)