For the reaction: 4PH3 + 802 → P4O10 + 6H2O a. If 45 g of PH3 and 35 g O2 were available, identify the limiting and excess reagents. b. If the experimental yield of P4O10 is 24 g, calculate the percentage yield.
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- Consider the series of reactions to synthesize the alum (KAl(SO4 )2 · xH2O(s)). If you start the synthesis with 1.00 g of Al, 40.0 mL of 1.50 M KOH, and 20.0 mL of 9.00 M H2SO4 , which of the three will be the limiting reagent?Consider the series of reactions to synthesize the alum (KAl(SO4 )2 · xH2O(s)) from the introduction. (a) Assuming an excess of the other reagents, from one mole of aluminum Al (s), how many moles of alum will be produced? (b) Assuming an excess of the other reagents, from one mole of potassium hydroxide KOH, how many moles of alum will be produced? (c) Assuming an excess of the other reagents, from one mole of sulfuric acid H2SO4 , how many moles of alum will be produced? (d) If you start the synthesis with 1.00 g of Al, 40.0 mL of 1.50 M KOH, and 20.0 mL of 9.00 M H2SO4 , which of the three will be the limiting reagent? (e) Assuming that the product is anhydrous (that there are no waters of hydration), calculate the theoretical yield of alum, in grams, based on the amounts of reagents in part (d). 3. Consider the nickel salt: (NH4 )2Ni(SO4 )2 ·y H2O (Ammonium Nickel Sulfate Hydrate), where y is the number of coordinated waters. (a) Assuming that the product is anhydrous (y = 0),…The mercury in a 1.1113-g sample was precipitated with an excess of paraperiodic acid, H5IO6: 5Hg2+ + 2H5IO6 → Hg5(IO6)2 + 10H+ The precipitate was filtered, washed free of precipitating agent, dried, and weighed, and 0.4733 g was recovered. Calculate the percentage of Hg2Cl2 in the sample. _______%
- 6 calculate % yield for the reaction below: Recrystallized product weight = 0.2303 gramsGive the theoretical yield, in moles, of CO2 from the reaction of 0.60 moles of C8H18 with 40.00 moles of O2. 2C8H18+25O2 ---> 16CO2 + 18H2OMass 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?
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- Chromium sulfate is an inorganic compound used as a tanning agent for leather manufacturing. It is produced after the reduction of potassium dichromate. The potassium dichromate is treated with 80% aqueous sulfuric acid solution and 20% excess H2O2 by the following reaction: K2Cr2O7 + 4 H2SO4 + 3 H2O2 à K2SO4 + Cr2(SO4)3 + 7 H2O + 3 O2 Analysis of the product mixture shows 40.21% Cr2(SO4)3, 4.11 % K2Cr2O7, 23.21% H2O. On a basis of 1000 kg of product mixture, determine the following: (a) Limiting reactant and excess reactants (b) % Excess of ER (c) Degree of completion (d) Illustrate The Law of Conservation of Mass (e) Complete mass composition of product mixtureWhat will be the mass of MnO2 that is needed to have an outcome of 25g Cl2? MnO2+4HCl->2H2O+Cl2 a. 70.30g b. 30.70g c. 37.0g d. 73.0gHow do we calculate the yield -- based on the following: Isopentyl alcohol (MW=88.2; d=0.813). Isopentyl acetate (MW=130.19; d=0.9 g/mL). Any assistance and breakdown of steps would be helpful. thanks!