Pennies minted after 1982 are mostly zinc (97.5%) with a copper cover. If a post-1982 penny is dissolved in a small amount of nitric acid, the copper coating reacts as in Problem 4.52, and the exposed zinc reacts according to the following equation: For a penny that weighs 2.482 g, what is the molarity of the Zn(NO3)2 if the resultant solution is diluted to 250.0 mL with water?

Video transcript

Hi everyone here we have a question telling us, the solution is created by dissolving 0.82 g of potassium hydroxide in 17.3 mL of 0.75 moller aluminum sulfate. And the question is how much aluminum and malaria t is left in the solution. So first we need a balanced equation. So we know we're starting off with aluminum sulfate and potassium hydroxide. Now we have to find out what we're forming. So aluminum is going to go with the hydroxide and we're going to use the criss cross method. So these will become our substitute parents. So we will have aluminum hydroxide and potassium has a plus one charge and it's going to form with sulfate which has a charge of two minus. So again we're gonna crisscross and we'll get potassium to sulfate and now we need to balance this. So if we look on our reactant side, we notice that we have two aluminums, we only have one on our product side. So we're going to put two there to fix that and now we have three hydroxide on our product side and only one on our reactant side. So or we have six hydroxide. So we're gonna put a six there And now we have two potassium on our product side and six on our reactive side. So we're going to put a three there and now we have our balanced equation. So now we can calculate our amount of reactant. So we have 0.82 g of potassium hydroxide Times one mole of potassium hydroxide divided by potassium hydroxide molar mass, which is found on the periodic table and is 56. g of potassium hydroxide. And our g are going to cancel out. And that's going to equal 0. moles of potassium hydroxide. And we're going to now calculate the moles of aluminum sulfate. So we have 17.3 ml and we're going to change that to leaders. So times one liter over 1000 ml Times 0.75 moles of aluminum sulfate over one liter of aluminum sulfate And that equals 0. moles of aluminum sulfate. If we look at our dimensional analysis, our middle leaders cancel out our leaders cancel out and we are left with our moles, which is what we want. So now we need to calculate our limiting re agent. So we have 0. moles of potassium hydroxide, times two moles of aluminum hydroxide over six malls, a potassium hydroxide. And that is found in our balanced equation. So that equals 0. moles of aluminum hydroxide. And now we're going to do the same thing with aluminum sulfate. So we have 0. moles of aluminum sulfate. And we're going to multiply that by two moles of aluminum hydroxide over one mole of aluminum sulfate. And that equals 0. moles of aluminum hydroxide. So because we get less aluminum hydroxide when we use our moles of potassium hydroxide, that means it is our limiting re agent are limiting reactant. So now we need to find the amount of aluminum sulfate we have remaining. So our aluminum sulfate consumed is 0. moles of aluminum. I dropped side times one mole of aluminum sulfate over two moles of aluminum hydroxide And that equals 0. 539 moles. So our remaining Is 0.012975 moles that we started out with minus the consumed. That we just calculated. So 0. moles. And that equals 0. 05391 moles. Now, we need to calculate our aluminum remaining. So we have our 0. moles, aluminum sulfate, times two moles of aluminum over one mole of aluminum sulfate. Because there's two moles in aluminum sulfate And that equals 0. moles of aluminum. And now we need to calculate that in polarity and polarity equals moles over leaders. So 0. moles over zero point 0173 ml equals 1.218 Polarity. And that is our final answer. Thank you for watching. Bye