Ans 1 :
The balanced reaction is given as :
2Na3PO4 + 3Ni(NO3)2 = Ni3(PO4)2 + 6NaNO3
Mol sodium phosphate = mass / molar mass
= 12.05 g / 163.94 g/mol = 0.0735 mol
Mol nickel (II) nitrate = molarity x volume (L)
= 0.500 M x 0.175 L
= 0.0875 mol
2 mol sodium phosphate needs 3 mol nickel (II) nitrate
so 0.0735 mol sodium phosphate will need : (0.0735 mol x 3 mol) /2 mol = 0.110 mol nickel (II) nitrate
but here only 0.0875 mol of nickel (II) nitrate is present , so it is the limiting reactant
so number of mol of nickel (II) phosphate formed will be : 0.0875 mol / 3 = 0.029 mol
mass of nickel (II) phosphate = mol x molar mass
= 0.029 mol x 366 g/mol
= 10.68 grams
Mol Na3PO4 utilised = (0.0875 mol x 2 mol) / 3 mol = 0.0583 mol
mol Na3PO4 left excess = 0.0735 mol - 0.0583 mol = 0.0152 mol
so mol of PO43- unprecipitated = 0.0152 mol
concentration of PO43- = mol / volume (L)
= 0.0152 mol / 0.175 L
= 0.0868 M
1. Balance the reaction below and determine how many grams of nickel(II) phosphate will be formed...
1. Balance the reaction below and determine how many grams of nickel(II) phosphate will be formed when sodium phosphate (12.05g) is mixed with nickel(II) nitrate (175 ml, 0.500 M). After the reaction is complete, what is the concentration of the unprecipitated ion (either PO or Ni²+). Na,PO + Ni(NO3)2 - Ni,(PO),+ + NANO,
Balance the reaction below and determine how many grams of nickel(II) phosphate will be formed when sodium phosphate (12.05g) is mixed with nickel(II) nitrate (175 mL, 0.500 M). After the reaction is complete, what is the concentration of the unprecipitated ion (either POX or Ni2+). Na3PO4 + Ni(NO3)2 — Ni3(PO4)2+ + NaNO3
how many moles of Lead(II) chloride are formed in the reaction? how many grams of lead (II) chloride are formed in the reaction? Procron 823 2. 25.0 mL of 0.150 M aqueous hydrochloric acid is mixed with 1.00 g of solid lead(II) nitrate dissolved in water. a.) Write balanced chemical equations for the process that occurs when these two solutions are mixed. Indicate the physical state of each reactant and product. The Thelin & PRETONOOD
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...
A sample of 1.45 g of lead (II) nitrate is mixed with 129 mL of 0.100M sodium sulfate solution. Part A: What is the concentration of SO42- ion that remains in solution after the reaction is complete? Part B: What is the concentration of Na+ ion that remains in solution after the reaction is complete? Part C: What is the concentration of NO3- ion that remains in solution after the reaction is complete?
For the reaction CaCO, + 2 NaNO Na,CO, + Ca(NO,)2 how many grams of calcium nitrate, Ca(NO3)2, are needed to react completely with 53.7 g of sodium carbonate, Na CO,? mass
Consider the following chemical reaction. Mg(s)+Ni(NO3)2(aq)⟶Ni(s)+Mg(NO3)2(aq) Which statement is true for the reaction? The nickel(II) ion is oxidized. The nitrate ion is the reducing agent. The nickel(II) ion is the oxidizing agent. Metallic magnesium is reduced.
How many grams of copper(II) nitrate are needed to produce 29.6 g copper(II) phosphate in the presence of excess sodium phosphate?
2. What will you observe when you have a positive test for: (a) bicarbonate anion(HCO3-)? • Bubbles Dobre lieT like gas o colorless (b) phosphate anion(PO43-)? 1. A student studied some reactions involving Pb2+ ion, copper(II) ion (Cu2+), and nickel(II)ion (Ni2+). The student first observed the behavior of individual solutions of Pb(NO3)2, copper(II) nitrate, Cu(NO3)2, nickel(II) nitrate, Ni(NO3)2, interacting individually with 1M HCl solution and then individually with 1M sodium hydroxide solution (NaOH). The student also added an aqueous ammonia...
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