I’m having some trouble with a couple of CHEMISTRY essay questions. Help would be greatly appreciated!?

How many grams of a stock solution that is 92.5 percent H2SO4 by mass would be needed to make 250 grams of a 35.0 percent by mass solution? Show all of the work needed to solve this problem.ANDHow many grams of iron metal do you expect to be produced when 305 grams of a 75.5 percent by mass iron(II) nitrate solution reacts with excess aluminum metal? Show all of the work needed to solve this problem.



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One Response to “I’m having some trouble with a couple of CHEMISTRY essay questions. Help would be greatly appreciated!?”

  1. preconversational says:

    In this problem, we will use the percentages as the concentrations and the masses as volume (where mass represents the volume of the liquid that will weigh that much). NB: Knowing the density and volume of a liquid it is easy to get the mass.M1V1=M2V2(i)M1=92.5 %; V1=?; V2=250 g; M2=35.0 %V1=(M2V2)/M1 = (250*35.0)/92.5 = 94.6 gSo you need the corresponding volume of 94.6 g of H2SO4 of a 92.5 % stock solution and and then add 155.4 ml of water (note 1 ml of H2O = 1 g since the density of water is 1 g/ml)(i) 3Fe(NO3)2 + 2Al (s) —> 2Al(NO3)3 + 3Fe (s)from the stoichiometric equation, 3 moles of iron nitrate reacts to give an equal (3) mole of iron metalAssume you had 305 g of 100 % solution of iron nitratemoles of Fe(NO3)2 = mass/MW = 305/179.8 = 1.7 molesMoles of Fe solid produced = 1.7 moles toomass of Fe = moles*MW = 1.7*55.8 = 94.1 gNow, if 305 g of 100 % Fe(NO3)2 = 94.1 g of Fe solid (produced) 305 g of 75.5% Fe(NO3)2 = 75.5 % of 94.1 g of Fe solid = 71.2 g of FeYou can solve this differently but I decided to word out the solution so you can follow along.