1 ) What volume of 0.150 M hydrochloric acid reacts with excess lead (II) nitrate solution...

1 ) What volume of 0.150 M hydrochloric acid reacts with excess lead (II) nitrate solution in order to yield 1.88 g lead ( II) chloride precipitate?

2) Calculate the boiling point and the freezing point of 50.0 g of AlCl3 dissolved in 400g of water, note that AlCl3 produces four particles

3) If 45.5 mL of 0.150 M sodium sulfate solution reacts with 50.0 mL of 0.175 M aqueous barium nitrate, what is the mass of Ba SO4 precipitate

Expert Solution

queen_honey_blossom answered 10 months ago

10.00g lead (II) nitrate reacts with 10.00g sodium iodide in aqueous solution to produce solid lead...

10.00g lead (II) nitrate reacts with 10.00g sodium iodide in aqueous solution to produce solid lead (II) iodide and aqueous sodium nitrate. What mass of NaI remains? I have the answer, and the process of using stoichiometry to find it. I just do not understand the logic of the stoichiometry and why you use it to find the NaI consumed. How is do you know it is "consumed" based off of the stoichiometry you do?

A solution contains barium nitrate and lead (II) nitrate. What substance could be added to the...

A solution contains barium nitrate and lead (II) nitrate. What substance could be added to the solution to precipitate the lead (II) ions, but leave the barium ions in solution? A) ammonium carbonate B) no such substance exists C) beryllium sulfate D) silver nitrate E) strontium bromide

Consider mixing an excess of lead(II) nitrate (aq) with 200.0 mL of 0.400 M sodium chloride....

Consider mixing an excess of lead(II) nitrate (aq) with 200.0 mL of 0.400 M sodium chloride. Determine the mass of solid formed, assuming a complete reaction What volume (in mL) of 0.800 M lead(II) nitrate must you add to make sure you make the mass of product you calculated in problem 3? A 0.1044 g sample of the salt MCl was dissolved in water (making it in the aqueous phase). An excess of AgNO3 (aq) was added, precipitating 0.0889 g...

What volume (L) of 0.986 M hydrochloric acid solution neutralizes 91.4 mL of 0.196 M sodium...

What volume (L) of 0.986 M hydrochloric acid solution neutralizes 91.4 mL of 0.196 M sodium hydroxide solution ? What volume (L) of 0.288 M potassium hydroxide solution would just neutralize 35.7 mL of 0.279 M H2SO4 solution? Write the balanced total ionic equation for the same reaction. Enter your answer as the sum of the coefficients. For the reaction HI + Pb(NO3)2 → ,write the balanced formula equation. Enter your answer as the sum of the coefficients. ALL MUST...

What is the molar concentration of a lead nitrate solution that is 18.0% (m/m) by mass...

What is the molar concentration of a lead nitrate solution that is 18.0% (m/m) by mass lead nitrate. Molar Mass[Pb(NO3)2] = 331.22 g/mol; Density[Pb(NO3)2] = 1.18 g/mL.

a sample of magnesium metal reacts completely with an excess of hydrochloric acid: MG(s) + 2...

a sample of magnesium metal reacts completely with an excess of hydrochloric acid: MG(s) + 2 HCl (aq) = MgCl2(aq) + H2(g). The hydrogen gas produced is collected over water at 28.0 C. the volume of the gas is 46 L and the total pressure in the gas collection vessel is 1.01 atm. calculate the mass of magnesium metal that reacted in grams (vapor pressure of water at 28.0 C = 28.2 mmHg)

What mass of ammonium nitrate must be added to 350 mL of a 0.150 M solution...

What mass of ammonium nitrate must be added to 350 mL of a 0.150 M solution of ammonia to give a buffer having a pH of 9.00. (kb(NH3)=1.8x10^-5). please show each step in detail.

Potassium iodide, Kl, reacts with lead(II) nitrate according to the following chemical equation.

Potassium iodide, Kl, reacts with lead(II) nitrate according to the following chemical equation. 2 Kl (aq) + Pb(NO3)2 (aq) → 2 KNO3 (aq) + Pbl2 (s) If 32.31 mL of 0.383 M Kl reacts with excess Pb(NO3)2 , what mass of Pbl forms? The molar mass of Pbl2 is 461.01 g/mol.

when a solution of lead(ii) nitrate is mixed with a solution of potassium chromate, a yellow...

when a solution of lead(ii) nitrate is mixed with a solution of potassium chromate, a yellow precipitate forms according to the equation: Pb(NO3)2 (aq) + K2CrO4 (aq) -> 2 KNO3 (aq) + PbCrO4. Volume of 0.105 M lead(ii) nitrate react with 100.0 ml of 0.120 M potassium chromate. What mass of PbCrO4 will be formed?

An aqueous solution containing 7.22g of lead (II ) nitrate is added to an aqueous solution...

An aqueous solution containing 7.22g of lead (II ) nitrate is added to an aqueous solution containing 6.02g of potassium chloride. Enter the balanced chemical equation for this reaction. Be sure to include all physical states.What is the limiting reactant?The % yield for the reaction is 84.1%, how many grams of precipitate were recovered? How many grams of the excess reactant remain?

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