What is the percent yield of the solid product when 18.37 g of iron(III) nitrate reacts in solution with excess sodium phosphate and 3.526 g of the precipitate is experimentally obtained? Fe(NO3)3(aq) + Na3PO4(aq) --> FePO4(s) + NaNO3(aq) [unbalanced]
What is the percent yield of the solid product when 18.37 g of iron(III) nitrate reacts...
99 Marks: 1 What is the mass (in grams) of the solid product when 6.83 g of iron(II) nitrate reacts in solution with excess sodium hydroxide? Fe(NO3)2(aq) + NaOH(aq) --> Fe(OH)2(s) + NaNO3(aq) [unbalanced]
The metathesis reaction between iron (III) nitrate and sodium sulfide is shown below 2 Fe(NO3)3(aq) + 3 Na2S (aq) ➝ 6 NaNO3(aq) + Fe2S3(s) How many grams of solid iron (III) sulfide can be produced by the reaction of 250.0 ml of 0.400 M iron (III) nitrate solution with 350.0 ml of 0.250 M sodium sulfide solution? Determine the final concentration of each ion in solution at the end of the precipitation reaction. Na+ NO3– Fe+3 S–2
Silver nitrate, AgNOs, reacts with iron(III) chloride, FeCl3,to give silver chloride, AgCl, and iron(III) nitrate, Fe(NO3). A solution containing 18.0 g of AgNO, was mixed with a solution containing 32.4 g of FeCls. How many grams of which reactant remains after the reaction is over?
When 84.8 g of iron (III)oxide reacts with excess CO in the laboratory, 54.3 g of iron is isolated. Fe2O3 + 3 CO = 2 Fe + 3 CO2 What is the actual yield, the theoretical yield, and the percent yield?
Given that 24.0 mL of 0.170 M sodium iodide reacts with 0.209 M mercury (II) nitrate solution according to the unbalanced equation Hg(NO3)2(aq) + Nal(aq) → Hgla(s) + NaNO3(aq) a) What volume of Hg(NO3)2 is required for complete precipitation of Hgl2? c) What is the mass of Hgla precipitate?
When a 5.12-g sample of solid sodium nitrate dissolves in 31.1 g of water in a coffee-cup calorimeter (see above figure) the temperature falls from 25.00 oC to 16.37 oC. Calculate H in kJ/mol NaNO3 for the solution process. NaNO3(s) Na+(aq) + NO3-(aq) The specific heat of water is 4.18 J/g-K. We were unable to transcribe this imageWe were unable to transcribe this imagethermometer stirrer coffee cups containing reaction mixture When a 5.12-g sample of solid sodium nitrate dissolves in...
4, Question Details Solid sodium phosphate is added to an aqueous solution of manganese(II) nitrate in a series of experiments. The salts react to form a precipitate according to the chemical equation 2 Na3PO4(aq) 3 Mn(NO3)2(aq)Mn3(PO4)2(s) 6 NaNO3(aq) In each experiment, the precipitate was collected, dried, and weighed Exp 1 Exp 2 Exp 3 Exp 4 Exp 5 Exp 6 Volume of manganese) 487.6 m nitrate Mass of sodium phosphate g Mass of precipitate 377.6 ml 323.7 mL 232.4 mL...
The mole fraction of iron(III) nitrate, Fe(NO3)3, in an aqueous solution is 1.41×10-2. The percent by mass of iron(III) nitrate in the solution is -----%.
If 2.50 g of aluminum nitrate (Al(NO3)3) reacts to form 1.83 g of solid aluminum sulfate, what is the percent yield? 2Al(NO3)3(aq) + 3(NH4)2SO4(aq) → Al2(SO4)3(s) + 6NH4NO3(aq)
Fill in the Blanks A solution containing 58.0 g of mercury(II) nitrate is allowed to react completely with a solution containing 16.642 g of sodium sulfate according to the equation below: Hg(NO3)2(aq) + Na2SO4(aq) + 2NaNO3(aq) + HgSO4(s) How many grams of solid precipitate will be formed if the reaction has a 100% yield? Solid precipitate grams How many grams of the excess reagent will remain ilter the Excess reagent remaining grams If the reaction has only an 80% yield,...