Suppose you have 3.00 g of powdered zinc metal, 3.00g of powdered silver metal and 500.0 mL of a 0.2 M copper(II) nitrate solution. (d) What is the molarity of Cu2+ ions in the resulting solution?

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Welcome back everyone. We need to consider the following substances. five g of powdered platinum and five g of powdered chromium metal, identify the metal that will react with 650 mL of a 6500.500 moller nickel three nitrate solution, assuming that the reaction goes to completion, calculate the polarity of nickel two plus ions remaining after the reaction is complete. So beginning with part one of our question, we want to recall our activity series in our textbooks in which we would recall that the metals listed towards the top are going to be our most reactive or most prone to oxidation, whereas the metals listed towards the bottom are going to be the least reactive and least prone to oxidation. And because our prompt compares chromium metal and platinum, recall that on our activity series, chromium is listed towards the top and what separates these two is nickel in between. And so based on this ordering, we would say that chromium is going to be the most prone to oxidation or most will say most reactive to oxidation. Platinum would be the least. But because chromium is the most prone to oxidation, we would say that therefore our chromium metal reacts with our nickel to nitrate solution. And so this would be our first answer. And now we just need to verify this with a reaction. So we have solid chromium reacting with nickel to nitrate and in this redox reaction we would form the falling product which is chromium three nitrate as well as solid nickel. And we can understand that chromium is oxidized here because it goes from its standard state having an oxidation state of zero being a medal to on the product side, chromium has an oxidation state of now plus three. When you solve for X where X is chromium. And so we need to now make sure that this is balanced. So we'll place the following coefficients to balance this equation out. So with these added coefficients, we can confirm that this equation is now balanced and written correctly. Now our next step is to determine our limiting reactant and because our final answer is in concern of nickel two plus ions, we need to figure out our moles of nickel solid produced from both of our reactant. So beginning with nickel solid produced from chromium. Since we know chromium is what is reacting. Were given a mass of five g of chromium in which we want to multiply by the molar mass to go from grams of chromium in the denominator, two moles of chromium in the numerator and our molar mass of chromium from our periodic table will find is 51.9961 g of chromium equivalent to one mole, canceling out grams of chromium. We're then going to move to our next factor which is to go from our multiple ratio between moles of chromium. Our reactant, two moles of our product solid nickel and we don't really need to put in the labeled er but for molds of chromium two moles of nickel. We can see that from our bounced equation we have a 2 to 3 molar ratio. So that is specifically two moles of chromium for three moles of nickel canceling out. Now moles of chromium we're left with molds of our product nickel produced in which we will find our molds equal to 0.1442 moles of Nickel made from our chromium reactant. Now we need to find or compare this two moles of nickels produced from our second reactant, nickel to nitrate. And so beginning with our mass or rather when we look at the prompt, we're actually not given a mass for nickel to nitrate. Were given a volume and Mueller concentration. And so because we see that our molar concentration recall so recall that molar itty is interpreted as moles per liter. And if we're given a volume in milliliters we can just convert this to leaders to use in our story geometry to get two moles of our nickel produced from nickel to nitrate. And so beginning with the conversion of our volume we have 650 mL of nickel to nitrate which will convert from middle leaders in the denominator to leaders in the numerator recall that our prefix milli tells us we have 10 to the negative third of our base unit. Leader canceling out male leaders. We can now go from leaders to our Mueller concentration of nickel to nitrate. Which we are going to interpret as moles of nickel to nitrate per leader of solution of nickel to nitrate in the denominator. And so we would plug in our concentration as 0.500 moles per liter of solution. So now canceling out leaders, we're at moles of nickel to nitrate our reactant. And now we can use the molar ratio between moles of nickel to nitrate in the denominator. And just so everything is visible. I'm going to move it over And then in our numerator, we have moles of solid nickel. So, plugging in our molar ratio, we can see that from our prompt, we have three moles of nickel to nitrate for three moles of solid nickel produced. So we have a 3 to 3 molar ratio. So now canceling out moles of nickel to nitrate. We're left with moles of our product nickel. And what we'll find is that we have our nickel produced from nickel to nitrate being 30.3 25 moles of nickel. And so we can see that .144, 2 moles of Nickel is less than . moles of Nickel. And so because our reactant chromium produced a smaller amount, we would say therefore chromium is our limiting reactant. And so we need to figure out how much of our nickel to nitrate. We have remaining. So something to note is that because we have a 3-3, which is really just a 1-1 molar ratio between our moles of nickel and moles of nickel to nitrate. If we already know how many moles of nickel made from nickel to nitrate being 0.3 25 moles of nickel. If we already know that. And we know the ratio between the reactant, nickel to nitrate and nickel is the same. We can understand that this 250.3 25 moles of nickel are moles of nickel are going to be therefore equal to our moles of nickel to nitrate. And so that means that we have . moles of nickel to nitrate initially. So we don't even have to do that calculation since we understand this. And so now we just need to figure out how much of our nickel to nitrate reacts with chromium. And so for our nickel to nitrate that reacts, We're going to begin with our mass of chromium given as five g of chromium From the prompt and sorry, this is five. And we're going to multiply by the ratio to go from grams of chromium two moles of chromium. So again, utilizing the molar mass where we have 51.9961 g of chromium from our periodic table equivalent to one mole of chromium, canceling out my grams of chromium, we're now going to go from moles of chromium In the denominator, two moles of Nickel to nitrate in the numerator. And so plugging in our molar ratio from our bounced equation above, we have a 2 to 3 molar ratio. So in our denominator we have two moles of chromium for three moles of nickel to nitrate that react. So canceling out moles of chromium. We're left with moles of nickel to nitrate, which tells us our nickel to nitrate that reacts is equal to a value of point 1442 moles of nickel to night trait that reacts. And so now with this part solved, we can finally figure out how much we have in excess of nickel to nitrate and so how much remains. So our nickel to nitrate that remains is just going to be the difference of what we have initially, which we determined as .325 moles of nickel to nitrate initially Subtracted from what we know reacts of it being what we solved above as .1442 moles of nickel to nitrate. And so this difference will give us our Nickel three nitrate remaining or what we have in excess equal to 0.1808 moles that remain. So now with this amount figured out because for our final answer, we need to figure out our polarity of nickel two plus ions. We're going to begin with our moles of our nickel to nitrate, which we just found that remain in excess being 20. moles of nickel to nitrate. And we want to multiply by a geometric ratio where we go from moles of nickel to nitrate, two moles of our Nickel two plus ion that make up this compound. And we can see that we just have one mole of nickel two plus making up our compound of one mole of nickel three nitrate. And so we would have a 1 to 1 molar ratio. So canceling out moles of nickel to nitrate were left with moles of nickel two plus ion equal to from this product we have a value equal to 0.1808 moles of our nickel to Pluskat ion. And now with our moles of nickel two plus catalon we can find its polarity by taking the molds that we have 20.1808 moles of nickel two plus which we just solved for. And we want to divide by our volume because we called that polarity is in terms of moles per liter. And so we're given that volume from the prompt As 650 male leaders of our nickel three nitrate solution. But we need to convert this to leaders. So multiplying by our conversion factor to go from mill leaders in the denominator to leaders in the numerator are prefix milli tells us that we have 10 to the negative third of our base unit. Leader canceling out. Male leaders were left with leaders in the denominator and we have moles per liter as our final units for more clarity. And this will simplify to our molar concentration of our nickel two plus ion being 20.278 moller of our Nickel two plus ions remaining after the reaction is complete. And so our final answer is this molar itty of nickel two plus ions, which corresponds to choice be in the multiple choice as the correct answer. So I hope that everything our view was clear and let us know if you have any questions.

Suppose you have 3.00 g of powdered zinc metal, 3.00g of powdered silver metal and 500.0 mL of a 0.2 M copper(II) nitrate solution. (d) What is the molarity of Cu2+ ions in the resulting solution?

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Key Concepts

Molarity

Stoichiometry

Copper(II) Nitrate Dissociation