A sample of impure tin of mass 0.535 g is dissolved in strong acid to give a solution of Sn2+. The solution is then titrated with a 0.0448 M solution of NO3-, which is reduced to NO(g). The equivalence point is reached upon the addition of 0.0344 L of the NO3- solution. Find the percent by mass of tin in the original sample, assuming that it contains no other reducing agents.

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everyone in this example, we have a solution of lead two plus produced from dissolving 20.617 g of impure lead sample and strong acid. The solution needed 0.263 liters of 0.568 molar solution of nitrate to reach the equivalence point where nitrate is reduced two nitrogen monoxide gas. And we need to determine the mass percent of lead in the sample if there's no other reducing agent. So we're going to write out two half reactions from our first main reaction, which would be taking our lead two plus carry on where we would show that it's going to gain two electrons to form solid lead and sorry, that should be lead solid. And we would see that for the oxidation of solid lead according to our standard electrode potential table which we would find in our textbooks or online we have a value equal to negative 0.13V. Now, for our next reaction we have the nitrate and ion Which reacts with four moles of hydride And gains three electrons to form our product, which would be nitrogen monoxide gas And then two moles of water. Where we would see that for the oxidation of nitrogen monoxide, we have a self potential value equal to 0.96V. According to our text books. Now we need the amount of electrons to be the same for both of these reactions. So we're going to multiply this equation here by a factor of two. And then our first equation, we're going to multiply by a factor of three. And in doing so we would be able to now add up these reactions so that we now have two moles of nitrate Reacting with eight moles of hydride and three moles of lead solid, Which is going to produce two moles of nitrogen monoxide gas, Plus four moles of water in liquid form, Plus three moles of lead, two plus Catalon. And now that we have this fully balanced redox reaction, we want to make note of our ratio of our moles of lead to our moles of the nitrate an ion. And we would see that according to this equation we have a coefficient of two in front of nitrate and a coefficient of three in front of lead. And so this would mean that we have a 322 Molar ratio. So we're going to use this as a conversion factor to find our grams of lead Where from the prompt? We're going to take the volume of our solution which has a volume from the prompt of 0.0263 L. We're going to multiply this and convert from leaders into moles by recalling that polarity is represented by moles divided by leaders. And so we're going to take that polarity value from the prompt and we're going to plug it in as a conversion factor where we would say we have 0.0568 moles for one liter of solution. And this is specifically moles of our nitrate an ion. So now we can go ahead and cancel out leaders. And now we're going to focus on converting from moles of nitrate two moles of lead. Where as we stated, we have a Ratio of three moles of lead for two moles of nitrate. So our last step is to now cancel out moles of nitrate. And now we're going to go from moles of lead. two g of lead where we would recall from the periodic table that we have a molar mass for one mole of lead equal to 207.2 g of lead. So now we're able to cancel out moles of lead. And we're left with grams of lead as our unit here. And in our calculators we should get a mass of lead equal to 0.4643 g of lead. And so now we need to go ahead and calculate our percent mass of lead. So our mass percent of lead. And we should recall that this is going to equal our mass or our grams of lead rather divided by the grams of our sample, Multiplied by 100%. And so what we're going to get here is our grams of lead above. We just calculated 0.4643 g of lead. And then according to the prompt, our grams of our sample is equal to 0. g. And so we're going to multiply this quotient by 100% and we should get a mass percent value for lead equal to 75.3%. So this here would be our final answer as our mass percent of lead to complete this example. I hope that you have or I hope that everything I explained was clear. And if you have any questions, just leave them down below. Otherwise, I will see everyone in the next practice video.

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Key Concepts

Stoichiometry

Titration

Percent by Mass