Stoichiometry, Limiting Reactants, Molarity?
First lesson : Never trust your friend if he says you balanced the equation incorrectly. Second lesson : Don't ask such easy and questions, try them yourself. Third lesson : There is only one person, i.e. me who can tell the correct answer: The correct equation is: Pb(NO3)2 + 2NaI -------------> PbI2 + 2NaNO3 Moles = Molarity*volume in liter moles of lead II nitrate = 0.25*0.10 = 0.025 moles moles of sodium iodide = 0.45*0.20 = 0.09 moles From above equation, we can see that 1 mole of Pb(NO3)2 reacts with 2 moles of NaI to produce 1 mole of PbI2, so 0.025 moles of Pb(NO3)2 will react with 0.025*2 = 0.05 moles of NaI. But NaI is 0.09 moles, i.e. 0.04 moles extra, hence Pb(NO3)2 is limiting and NaI is in excess. PbI2 is the precipitate and has molecular mass = 461g mass of PbI2 = moles*461 moles of PbI2 = moles of Pb(NO3)2 = 0.025 so, mass of PbI2 = 0.025*461 = 11.525g I hope you can now solve the other parts.
What is the limiting reactant with mL and molarity?
Before you can determine the limiting reactant, you MUST write the equation for the reaction. 4OCl + S2O3 = 4Cl + SO3 + SO2 + O2 mole ratio of OCl : S2O3 is 4 : 1 At the same molar concentration, the OCl will have to be 4x the volume of the S2O3. You have 45 ml vs 5 ml, so it is 9 times the volume, thus making the S2O3 limiting.
Solve for the Molarity using the Limiting Reactant?
Suppose 2.94g of lead(II) nitrate is dissolved in 300.mL of a 39.0mM aqueous solution of ammonium sulfate. Calculate the final molarity of lead(II) cation in the solution. You can assume the volume of the solution doesn't change when the lead(II) nitrate is dissolved in it. Molar Mass of: Pb(NO3)2 = 331.2 g/mol (NH4)2SO4 = 132.14 g/mol Is this the balanced equation as well? Pb(NO3)2 + (NH4)2SO4 --> PbSO4 + 2NH4NO3
What is the molarity of excess reactant remaining in solution after precipitation?
Here is the full question from my textbook: What is the maximum mass, in grams, of AgCl that can be precipitated by mixing 50.0 mL of 0.025 M AgNO3 solution with 100.0 mL of 0.025 M NaCl solution? Which reactant is in excess? What is the concentration of the excess reactant remaining in solution after the AgCl has precipitated? Balanced equation: AgNO3(aq) + NaCl(aq) ---> AgCl(s) + NaNO3(aq) Since there are equal molarities and equal molar ratios, the larger volume is in excess. Therefore, NaCl is in excess. I use the limiting reactance, AgNO3, in determining maximum amount of precipitate. 50.0 mL AgNO3 * ( (0.025 mol AgNO3) / 1000 mL ) * ( 1 mol AgCl / 1 mol AgNO3 ) * ( (143.32 g AgCl) / 1 mol AgCl) = 0.179 g AgCl 0.179 g AgCl is the maximum. I know these answers are right (they are in the back of the book). However, I cannot get the last part of the question. I thought that 50.0 mL of AgNO3 reacted with 50.0 mL of NaCl. Since we start with 100.0 mL NaCl, we have 50.0 mL in excess. I saw this as a solution dilution problem and got: 100.0 ml * 0.025 M = 50.0 mL * x M solve for x. x M = 0.05 M However, the back of the book says that there is 0.0080 M NaCl left. Can someone explain this final step?
What is the limiting reactant in this experiment?
In this experiment, first 4.0 mL of methyl salicylate (HOC6H4COOCH3) with a density of 1.18 g/mL at 20 degrees Celsius is mixed with an aqueous solution of 40 mL of 6M (molarity or moles per liter) sodium hydroxide (NaOH). After these two chemicals are mixed, a white precipitate immediately forms, but this is not the desired product... The ionic compound in this precipitate is NaOC6H4COOCH3. The solution was boiled for 15 minutes until the precipitate dissolved, 50 mL of distilled water was added to the beaker, and then cautiously 50 mL of 8M (molarity) H2SO4 was added while stirring. The solution was filtered and excess water was removed; the resulting product was crystallized salicylic acid!! Now I am confused how to even begin with balancing the chemical formula of this experiment... The recorded mass of the salicylic acid was 2.934 g, and the mass of the methyl salicylate was 3.244g. I also calculated the mass of NaOH through the molarity to be 9.599 g of NaOH, but I'm not even sure if that is relevant... Please help me in finding the limiting reactant of this experiment and why it is the limiting reactant, thanks!
How do you find limiting reactants? It's been awhile for me...?
Please help me remember how to do these problems from Chem I....It's been awhile, and I really need help remembering. A student adds 20.0 mL of 3.00 M CaCl2 to 40.0 mL of 0.750 M K3PO4. Is CaCl2 or K3PO4 the limiting reactant? Also.. What is the molarity of the reactant that is not the limiting reactant and the molarity of KCl after the reaction occurs? (Assume any change in volume during the reaction is negligible). This is the reaction given: 3 CaCl2 (aq) + 2 K3PO4(aq) ----> Ca3(PO4)2 (s) + 6 KCl (aq) 1. What is the limiting reactant? 2. What is the molarity after the reacton for the reactant that is not the limiting reactant? 3. What is the molarity after the reacton for KCl? Any help would be greatly appreciated :)
20 ml of 0.1 molar HCl is reacted with NAOH of the same molarity until the pH of the solution is 4. How many ml of NAOH are still required to neutralize the solution completely?
HCl + NaOH —→ NaCl + H2OMolarity of HCl sol’n = 0.1 Mvol of HCl sol’n = 20 mLno. of milli mol of HCl = 2Molarity of NaOH sol’n = 0.1 Mlet vol of NaOH sol’n = x (in mL)no. of milli mol of NaOH = 0.1xSince the question asks us how much mL of NaOH is STILL needed for complete neutralization, it means that x < 20so, 0.1x < 2Hence, NaOH will act as the limiting reagentupon reaction,left over milli mol ofHCl = 2–0.1xNaOH = 0formed milli mol ofNaOH = 0.1xwater = 0.1xHCl, in sol’n exists as hydronium and chloride ion, somilli mol of H+ = 2–0.1xconc. of H+ ion in sol’n now, [H+] = (milli mol of H+ ion) / volume (in mL) of sol’nSo, [H+] = ( 2–0.1x ) / (20+x) ….(1)Since, now pH of sol’n is 4, [H+] = 0.0001 ….(2)Using (1) and (2), we get x = 19.96So, for complete neutralization, we need more 20 - 19.96 = 0.04 mL of 0.1 M NaOH sol’n
How do I find the limiting reactant, percent yield and theoretical yield for this?
1. Balanced equation should be: 2 KCl(aq) + Ba(NO3)2(aq) ---> BaCl2(s) + 2KNO3(aq) 2. Balanced equation shows we need twice as moles of KCl as Ba(NO3)2 - a 2:1 ratio. Lets see how much we have of each. moles = Molarity x Vol a.) KCl: 1.20-mol/L x 0.025-L = 0.03 mol b.) Ba(NO3)2: .900-mol/L x .015-L = 0.0135 mol x 2 = 0.027 c.) since the KCl we need is 2 x mol of Ba(NO3)2, we see we have slightly more than 2 times so Ba(NO3)2 is the limiting reagent. 3. The mass of BaCl2 that should form is: 0.0135 mol Ba(NO3)2 x (1 mol BaCl2 / 1 mol Ba(NO3)2) x ( 208.3- g BaCl2 / 1 mol BaCl2) = 2.81-g BaCl2 4. % yield = 2.45-g / 2.81-g x 100% = 87.2% yield
What is the molarity of a final solution after mixing 30 ml of 0.2M HCL and 20 ml of a 0.1M naoh solution?
So this is the Answer…..1}first write the chemical reactionNaOH+HCL—→Nacl+H20See this reaction is already balanced so no need ….2}Then use the Limiting reagent concept try to identify the limiting reagent and the excessive reagentTelling how to do thatcheck how many molarity is given multiply with volume then divide it by the cofficent{Here cofficent of NaOH is 1 and also HCL IS 1}}YOU WILL GET MOLES ASMOLES OF SOLUTE=MOLARITY multiply with VOLUMENaOH-2molHCL-6molHere moles of HCL IS GREATER THAN moles of NaOH so …NaOH- LIMITING REAGENTHCL-EXCESSIVE REAGENTthen/….if NaOH is the limting reagent all the moles will utilised and hence it becomes 0.SO FINAL Moles of HCL=moles of NaOH- intial moles of HCL=>6–2=4molFinal volume=30+20=50mlchange in litre=>1/20 molSo Molarity = 4/20= 1/5 M Ans……………