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NH3 (g) formed. Identify the limitingreagent in the production of NH3 in this situation


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please check if the answers for the following questions are correct :)

1.) Given the balanced equation, Fe2O3 + 3 CO -> 2 Fe + 3 CO2, what is the molar ratio of CO to CO2?3:33:13:21:32.) Given the reaction, H2 + Cl2 -> 2HCl, if 12.0 moles of HCl were made, then how many moles of H2 were used?24.0 moles12.0 moles6.0 moles3.0 moles3.) How many atoms are in 5.00 moles of water (H2O)?8.33 x 10−24 atoms6.02 x 1023 atoms3.00 x 1024 atoms9.03 x 1024 atomsQuestion 4A sample of 78 moles of a gas, at standard temperature and pressure, occupies approximately what volume?3.48 L6.02 L1,747 L6.02 x 1023 LQuestion 5How many moles of a gas are contained in 11.2 liters of a gas, under standard temperature and pressure conditions?22.4 moles11.2 moles1 mole0.5 molesQuestion 6Using the equation, 4Fe + 3O2 arrow 2Fe2O3, if 6 moles of oxygen and an excess of iron were available, how many moles of iron (III) oxide would be produced?3 moles4 moles5 moles6 molesQuestion 7When magnesium metal and an aqueous solution of hydrochloric acid combine, they produce an aqueous solution of magnesium chloride and hydrogen gas. Using the equation, Mg (s) + 2HCl (aq) arrow MgCl2 (aq) + H2 (g), if 48.6 g of Mg and 150.0 g of HCl are allowed to react, identify the limiting reagent.MgHClMgCl2H2Question 8When magnesium metal and an aqueous solution of hydrochloric acid combine, they produce an aqueous solution of magnesium chloride and hydrogen gas. Using the equation, Mg (s) + 2HCl (aq) -> MgCl2 (aq) + H2 (g), if 24.3 g of Mg and 75.0 g of HCl are allowed to react, calculate the mass of MgCl2 that is produced.75.0 g95.1 g99.3 g104 gQuestion 9In order to properly calculate which reactant is limiting, the amount of each reactant, in grams, must be compared.TrueFalseQuestion 10Red mercury (II) oxide decomposes to form mercury metal and oxygen gas according to the following equation: 2HgO (s) arrow 2Hg (l) + O2 (g). If 3.00 moles of HgO decompose to form 1.25 moles of O2 and 503 g of Hg, what is the percent yield of this reaction?41.6%62.5%83.3%96.9%Question 11Ammonia gas is formed from nitrogen gas and hydrogen gas, according to the following equation: N2 (g) + 3H2 (g) arrow 2NH3 (g). If 84.0 g of nitrogen gas is allowed to react with an excess of hydrogen gas to produce 85.0 g of ammonia, what is the percent yield of this reaction?42.2%65.0%70.3%83.3%Question 12In the manufacturing process of sulfuric acid, sulfur dioxide is reacted with oxygen to produce sulfur trioxide. Using the equation, 2SO2 (g) + O2 arrow 2SO3 (g), if 64.06g of sulfur dioxide is given an opportunity to react with an excess of oxygen to produce 75.00 g of sulfur trioxide, what is the percent yield of this reaction?46.83%60.25%75.55%93.68%Question 13The theoretical yield for a chemical reaction can not be calculated until the reaction is completed.TrueFalseQuestion 14Percent yield is calculated by the formula, (Actual Yield/Theoretical Yield) x 100%.TrueFalse
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1) This is determined by looking at the ratio of the stoichiometric coefficients of CO to CO2, which is 3:3.2) Looking at the equation, the ratio of the number of moles of H2 that reacts to that of HCl that forms is 1:2. Thus, the number of moles of H2 that were used= 1/2(number of moles of HCl that were formed)= 1/2(12.0)= 6.0 moles3) 1.00 mole of H2O= 6.02x10^23 H2O molecules5.00 moles of H2O= 5x6.02x10^23 H2O molecules= 3.01x10^24 H2O moleculesAn H2O molecule contains 3 atoms. Thus, the number of atoms in 3.01x10^24 H2O molecules= 3x3.01x10^24= 9.03x10^24 atoms4) 1 mole of a gas occupies 22.4L at STPThus, the volume occupied by 78.0 moles of a gas at STP= (78.0x22.4)L= 1747.2L= 1747L(answer correct to 3 significant figures)5) 22.4L of a gas contains 1 mole of a gas at STPThus, the number of moles contained by 11.2L of a gas at STP= 11.2/22.4= 0.5 moles6) Since Fe is in excess, O2 is the limiting reagent, implying that all of it reacts.The ratio of the number of moles of O2 that reacts to that of Fe2O3 that forms is 3:2.Thus, the number of moles of Fe2O3 that form= 2/3(number of moles of O2 that reacts)= 2/3(6.00)= 4.00 moles7) Relative Atomic Mass(RAM) of Mg= 2424.0g= 1 mole of Mg48.6g= 48.6/24.0= 2.025 moles of MgRelative Formula Mass(RFM) of HCl= (RAM of H)+(RAM of Cl)= 1.0+35.5= 36.536.5g= 1 mole of HCl150g= 150/36.5= 4.110 moles of HClAccording to the equation, the ratio of the number of moles of Mg required to that of HCl required is 1:2. Since there is less Mg than required, Mg is the limiting reagent.8) 24.0g= 1 mole of Mg24.3g= 24.3/24.0= 1.0125 moles of Mg36.5g= 1 mole of HCl75.0g= 75.0/36.5= 2.0548 moles of HClAgain, the ratio of the number of moles of Mg required to that of HCl required is 1:2. Since there is more HCl than required, Mg is the limiting reagent, implying that all of it reacts.The ratio of the number of moles of Mg that reacts to that of MgCl2 that forms is 1:1. Therefore, the number of moles of MgCl2 that forms= the number of moles of Mg that reacts= 1.0125Relative Formula Mass(RFM) of MgCl2= (RAM of Mg)+2(RAM of Cl)= 24+2(35.5)= 951 mole of MgCl2= 95.0g1.0125 moles of MgCl2= 95.0x1.0125= 96.1875g9) False. It is actually the number of moles of each reactant that must be compared.10) Relative Atomic Mass of Hg= 200.6200.6g= 1 mole of Hg503.0g= 503.0/200.6= 2.51 moles of HgThe expected number of moles of O2 produced= 1/2(3.00)= 1.50Percent yield= (actual yield/theoretical yield)x100%= (1.25/1.50)x100%= 83.3%Relative Molecular Mass(RMM) of N2= 2(RAM of N)= 2(14)= 2828.0g= 1 mole of N284.0g= 84.0/28.0= 3 moles of N2Relative Molecular Mass(RMM) of NH3= (RAM of N)+3(RAM of H)= 14+3(1)= 1717.0g= 1 mole of NH385.0g= 85.0/17.0= 5.00 moles of NH3The expected ratio of the number of moles of N2 that reacts to that of NH3 that forms is 1:2. Thus, the percent yield= (actual amount of NH3/expected amount of NH3)x100%= (5/6)x100%= 83.3%12) Relative Molecular Mass(RMM) of SO2= (RAM of S)+2(RAM of O)= 32+2(16)= 6464.00g= 1 mole of SO264.06g= 64.06/64.00= 1.0009 moles of SO2RMM of SO3= 32+3(16)= 8080.0g= 1 mole of SO375.0g= 75/80= 0.93750 moles of SO3The expected ratio of the number of moles of SO2 that reacts to that of SO3 that forms is 1:1. Thus, percent yield= (0.93750/1.0009)x100%= 93.68%13) False. The theoretical yield can actually be calculated before the experiment has commenced by determining the the expected ratio of the number of moles of the limiting reagent to that of the product.14) True.

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