Iron content in ores can be determined by a redox procedure in which the sample is first reduced with Sn2+, as in Problem 4.130, and then titrated with KMnO4 to oxidize the Fe2+ to Fe3+. The balanced equation is
What is the mass percent Fe in a 2.368 g sample if 48.39 mL of a 0.1116 M KMnO4 solution is needed to titrate the Fe3 + ?

Question

Iron content in ores can be determined by a redox procedure in which the sample is first reduced with Sn2+, as in Problem 4.130, and then titrated with KMnO4 to oxidize the Fe2+ to Fe3+. The balanced equation is 
What is the mass percent Fe in a 2.368 g sample if 48.39 mL of a 0.1116 M KMnO4 solution is needed to titrate the Fe3 + ?

Accepted Answer

Welcome back everyone. We're told that the amount of golden or can be determined by redox hydration. Gold is first oxidized from gold to gold three plus by adding a 1 to ratio volume by volume of nitric acid and hydrochloric acid. Then is precipitated using excess hydroquinone solution as shown below. So we have two moles of our gold canyon reacting with three moles of our hydroquinone to produce two moles of solid gold, six moles of hydrogen and three moles of our oh benzo qanon product. Then we have excess hydroquinone tie traded with Siri um sulfate in a back to hydration. We need to calculate the mass percent of gold in 0.4685 g. Or sample if the backed iteration or if after the titillation, it was found that 35.68 mL of 0.0 to moller hydroquinone was used to precipitate all of the gold three plus cat ion in the solution. So what we need to first begin with is finding that massive gold. We know from our periodic table that our molar mass of gold is equal to 196.97 g per mole. So we're going to find the mass of the gold in the ore by taking the volume. Used to precipitate our gold catalon in solution being 35.68 mL of our hydroquinone. So C six H 602. We're going to convert from mill leaders to leaders by recalling that are prefix milli tells us that we have 10 to the negative third power of our base unit leaders canceling out. Male leaders were now going to convert from that concentration of our hydroquinone which is given in the prompt as 0.25 moller. We want to recall that molar concentration can be interpreted as moles per liter. And so we're going to plug it in as 0.0 to moles per liter of our hydroquinone. So C six H 602 in both the numerator and denominator. So now we can cancel out our leaders of hydroquinone And now going from moles of hydroquinone. We want to get two moles of gold. So we're going to look at the molar ratio between our bounced reaction and we see that we have two moles of our gold cat eye on for three moles of our hydroquinone reactant. And so we're going to carry this below in the next line to multiply by that as a conversion factor where our three moles of our hydroquinone will go in the denominator and our two moles of our gold cat ion go in the numerator. Again this comes from our balanced reaction given in the prompt. Now we can cancel out our moles of hydroquinone and focus on getting two g of gold by and sorry, so just so that we're plugging in the proper factor in the numerator. We want to plug in our Gold solid product here. So we see we have two molds of that gold solid product. So we plug in two moles of solid gold and now we can go from moles of gold to gramps by recalling that we have a molar mass of 196.97 And sorry, this would go in our numerator because we want mass to be our final unit. So 196.97 g of gold from our periodic table for one mole of gold. And this allows us to cancel out our units of moles of gold. And what we're going to yield is our massive gold equal to 0.1171 g of gold. Now that we have our massive gold, we can find our final answer to get our mass percent of gold from the ore. So we would say that our mass percent is calculated by taking the mass of our gold divided by the mass of our your given in the prompt. And this is going to be multiplied by 100. So plugging in what we know, we have our mass of our gold and our numerator 1000.1171 g of gold divided by our massive r ore given in the prompt as 0. g of ore multiplied by 100. And we're going to yield a mass percent of 24.99 which will round to about 25%. And this would be our final answer as our mass percent of gold in our or so. I hope that everything I reviewed was clear. If you have any questions, please leave them down below, and I will see everyone in the next practice video.

Verified Solution

Key Concepts

Redox Reactions

Titration

Molarity and Mass Percent