Tin(IV) sulfide, Sns,, a yellow pigment, can be produced using the following reaction. SnBr.(aq) + 2...
Heted 15 out of 20 Suppose a student als 42.3 mL of a 0.412 M solution of SnBr, to 41.3 mL of a 0.121 M solution of Na,s. Identify the limiting reactant. Na,s NaBr SnBr, Sns, Calculate the theoretical yield of Sus, theoretical yield: The student recovers 0.243 g of Sns, Calculate the percent yield of Sns, that the student obtained. percent yield:
5. A student performs the following reaction using the procedure outlined in Part A of this experiment. Ba(NO3)2 (aq) + K2SO4 (aq) - BaSO4(s) + 2 KNO3(aq) The student adds 55.0 mL of 0.050 M K SO. solution to 50.0 ml of 0.050 M Ba(NO3)2 solution. The precipitate of Baso. is collected by gravity filtration and dried to a constant mass. The BaSO.is determined to have a mass of 0.49 g. (See section 3.9 in the textbook for a review...
A precipitation reaction occurs when 749 mL of 0.846 M Pb(NO3)2 reacts with 375 mL of 0.810 M KI, as shown by the following equation. Pb(NO3)2(aq) + 2 Kl(aq) – PbL,(s) + 2 KNO, (aq) Identify the limiting reactant. O Pb(NO3)2 ОРЫ, OKI O KNO, Calculate the theoretical yield of Pol, from the reaction. mass of Pble: Calculate the percent yield of Pbl, if 50.6 g of Pol, are formed experimentally. percent yield of Pbl Suppose 57.2 mL of a...
Lead ions can be precipitated from solution with KCl according to the following reaction: Pb2+(aq)+2KCl(aq) →PbCl2(s)+2K+(aq) When 28.5 g KCl is added to a solution containing 25.5 g Pb2+, a PbCl2 precipitate forms. The precipitate is filtered and dried and found to have a mass of 29.3 g. Determine the limiting reactant. Determine the theoretical yield of PbCl2. Determine the percent yield for the reaction.
Chlorine gas can be prepared in the laboratory by the reaction of hydrochloric acid with manganese(IV) oxide. 4 HCI(aq) + MnO2(s) MnClj(aq) + 2H20) +C2(g) A sample of 38.7 g MnO2 is added to a solution containing 41.3 g HC. What is the limiting reactant? O HCI O Mno, What is the theoretical yield of Cl2? theoretical yield: g Cl If the yield of the reaction is 78.9%, what is the actual yield of chlorine? actual yield: gC2 Chlorine gas...
Chlorine gas can be prepared in the laboratory by the reaction of hydrochloric acid with manganese(IV) oxide. 4HCl(aq)+MnO2(s)⟶MnCl2(aq)+2H2O(l)+Cl2(g) A sample of 37.9 g MnO2 is added to a solution containing 42.3 g HCl. What is the limiting reactant? MnO2 OR HCl What is the theoretical yield of Cl2? If the yield of the reaction is 79.1%, what is the actual yield of chlorine?
Consider the following reaction. MgCl2(aq)+2NaOH(aq)⟶Mg(OH)2(s)+2NaCl(aq) A 166.0 mL solution of 0.381 M MgCl2 reacts with a 47.33 mL solution of 0.568 M NaOH to produce Mg(OH)2 and NaCl. Identify the limiting reactant. NaOH Mg(OH)2 MgCl2 NaCl Caclulate the mass of Mg(OH)2 that can be produced. The actual mass of Mg(OH)2 isolated was 0.559 g. Calculate the percent yield of Mg(OH)2.
Chlorine gas can be prepared in the laboratory by the reaction of hydrochloric acid with manganese(IV) oxide. 4 HCl(aq) + MnO,(s) —MnCl(aq) + 2 H2O(l) + Cl2(g) A sample of 42.1 g Mno, is added to a solution containing 42.1 g HCI. What is the limiting reactant? MnO2 OHCI What is the theoretical yield of CI,? theoretical yield: CI, If the yield of the reaction is 77.5%, what is the actual yield of chlorine? actual yield: g CI Dimethyl oxalate,...
The following chemical reaction takes place in aqueous solution: Fe2(SO4), (aq)+6 KOH(aq) — 2 Fe(OH)3(s)+3 K, SO4(aq) Write the net ionic equation for this reaction. 0- Identifying the limiting reactant in a drawing of a mixture The drawing below shows a mixture of molecules: key carbon hydrogen nitrogen sulfur 000000 oxygen chlorine Suppose the following chemical reaction can take place in this mixture: CH,(9)+4S(s) → CS (9)+2 H, S(9) of which reactant are there the most initial moles? Enter its...
Pre-Laboratory Assignment 1. Briefly describe the hazards associated with us ing the following (1) concentrated NH, solution 3. We can prepare the yellow coordination compound potassium tris(oxalato rhodate(lll), KIRh(,0)), by reacting potassiumn hexachlororhodatelli), KIRCI. and potassium oxalate, K 0,04 (1) Balance the chemical equation for this reac- tion, shown below. K(RhCi (aq) + K C20 (aq) → K[Rh(C2O4)3 (s. yellow) + KCl(aq) (2) If we combine 1 mol of KIRhClel with 1 mol of K C204, which compound is the...