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Please answer the question "Number of dosage form and packaging that can be produced from stoichiometric calculation."
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- Given Active Ingredient: precipitated sulfur (ointment) Raw Materials: 500 g calcium polysulphide and 1.5 kg hydrochloric acid Actual Yield: 343.4g precipitated sulfur Formulation: 250 mg per jar Dosage form: Ointment packaging:100 jars per box Synthesis and Packaging (Need answer)- Balanced Chemical Equation:- % composition by mass of each compound:- Mass to Mass Stoichiometry Calculation:- Limiting Reagent:- Excess Reagent:- Amount (g) in excess: % Yield:- Number of dosage form and packaging that can be produced from stoichiometric solution:balance C6H12O6+CO2+H2O with explanationConsider a blend of mass 5g formed from 2.2 g of C12H24 and 2.8g of C998H1000 paraffin's:, What are Mn and Mw of the blend?
- Sample: Saline 0.900% (m/v), in sodium chloride (NaCl). Data: M.A. (g/mol): Na = 22.9898; K = 39.0983; Cr=51.9961; Ag = 107.8682; Cl = 35.453; N=14.0067; O = 15.9994. Material available in the laboratory's warehouse: Reagents: distilled water; standardized solution of silver nitrate (AgNO₃), at a concentration of 0.09980 mols/L; Potassium chromate solution (K₂CrO₄) 1%; ammoniacal ferric alum solution [Fe(NH₄)(SO₄)₂] and; nitrobenzene (or cooking oil, alternatively); 0.1000 mols/L potassium thiocyanate (KSCN) standard solution. Glassware: beakers of all sizes available on the market; 50, 100, 150, 200 and 250 ml Erlenmeyers; 10.00 and 25.00 mL volumetric pipettes; 15.00 mL burette; 30.00, 50.00 and 100.00 mL volumetric flasks Calculate the molarity of saline, in terms of the NaCl concentration (concentration in mols/L). (a) 0.900 mols/L (b) 5.84 mols/L (c) 0.154 mols/L (d) 0.0154 mols/LGivenRaw Materials: 500 g calcium polysulphide and 1.5 kg hydrochloric acid Actual Yield: 343.4g precipitated sulfurFormulation: 250 mg per jarDosage form: Ointment Packaging: 100 jars per boxWhereas the balanced chemical reaction, as given in the problem is: CaS5+2HCl => 4S + CaCl2 + H2SMy problem: Find the number of dosage form and packaging that can be produced from the stoichiometric calculation based on the result upon finding the Mass to Mass Stoichiometry Calculation. I have based my answer to the mass stoichiometry from this link: https://www.bartleby.com/questions-and-answers/given-raw-materials-500-g-calcium-polysulphide-and-1.5-kg-hydrochloric-acid-actual-yield-343.4g-prec/b1de1102-f27f-4c8d-9b4e-c7fe8e6104b6Thanks, by the way. But can you still help me find the number of dosage from and packaging?Mass of Na2CO3.H2O (g) = 2.12g (g) Mass of the CaCl2.2H2O (g) = 1.98g Mass of the top funnel + filter paper (g) = 15.85g Mass of top funnel + filter paper + CaCO3 collected (g) = 17.81g CaCl2 + Na2CO3 ==== CaCo3 + 2NaCl Theoretical yield in moles and grams? Moles of reagent in excess left unreacted? Mass of precipitate? Experimental yield? Percent yield?
- Citric acid, C6H8O7, a component of jams, jellies, and fruity soft drinks, is prepared industrially via fermentation of sucrose by the mold Aspergillus niger. The equation representing this reaction is C12H22O11 + H2O + 3O2 --> 2C6H8O7 + 4H2O Using a metric ton(1000kg) of sucrose the expected yield is 1122Kg. On average if only 1036kg is produced, what is the % yield for this reaction?Assume that 1.00 mL (1.02 g) of crude product mixture is obtained from the reactionbefore the washing steps, and the distribution constant of the product mixture in brine isK = (Cmixture/Cbrine) = 50.0. Assuming that after mixing the product mixture volume isstill 1.00 mL (i.e., the volume lost is small), and the volume of the first brine wash layeris 2.00 mL. What mass (grams) of product mixture is lost (dissolves) in the first brinewash? (See Technique 13.2, p 53.) Show work.42. The primary use of 1,2‐dichloroethane, ClCH2CH2Cl, is to make vinyl chloride, which is then converted into polyvinyl chloride (PVC) for many purposes, including plastic pipes. Balance the following equation, which describes the industrial reaction for producing 1,2‐dichloroethane. C2H4 + HCl + O2 → ClCH2CH2Cl + H2O 50. Gaseous propane and liquid diethyl ether mix to form a solution. Which of thesesubstances is the solute and which is the solvent? 51. Consider a solution of 10% liquid acetone and 90% liquid chloroform. Which of these substances is the solute and which is the solvent?
- You are supplied with 1150 kg ethanol (C2H5OH), together with methanol (CH3OH), potassium hydroxide (KOH) and ammonia (NH3). You are requested by a customer to produce methyl acetate (CH3COOCH3), potassium acetate (CH3COOK) and ammonium acetate (CH3COONH4) in mass forms. (a) Calculate the mole & mass of acetic acid (CH3COOH) converted from ethanol through excessive oxidation.Complete and balance ---Co2(CO3)3(s)+heat---> ---H2O(l)+---As2O5(s)---> ---NaClO3(s)+heat---> ---Mg(s)+---H2SO4(aq)--->About 200 grams of glucose was fermented with 20 grams of yeast inside a fermenting vessel for 2 weeks. Calculate the percent yield of ethyl alcohol in mL or %(v/v) if the actual yield of the fermentation set-up is at 84.50 mL alcohol. Assume the density of ethanol is at 0.789 g/mL. C6H12O6 + H2O --> 2C2H5OH + 2CO2 Molar Masses: Glucose: 180.156 g/mole; Ethanol: 46.07 g/mole Water: 18.00 g/mole Report your answers in two decimal places. Please include the unit. For example: 55.33%