The reaction of tungsten hexachloride (WCl₆) with bismuth gives hexatungsten dodecachloride (W₆Cl₁₂).

WCl₆ + Bi → W₆Cl₁₂ + BiCl₃ Unbalanced

(c) When 228 g of WCl₆ react with 175 g of Bi, how much W₆Cl₁₂ is formed based on the limiting reactant?

Video transcript

Hey everyone in this example, we need to go ahead and write the balanced chemical equation for our described precipitation reaction and calculate the amount of nickel to sulfide that forms. So we have an a quick solution of nickel to chloride as one of our reactant. So that's N. I. C. L. To reacting with a second solution of sodium sulfide. So we're given that formula as N. A. Two S. We're told we have a precipitation reaction. We want to recall that precipitation reactions are double replacement reactions. And so we should recall that in a double replacement reaction. We pair up the first and last ions to form a product and then we pair up the middle two ions to form a second product. And so what we're going to have is our first product where we pair up nickel with sulfur. We should recall that nickel has a plus one charge, whereas sulfur has a minus one charge. So we would form N. I. S. As our product. And then for our second product, when we combine Chlorine with sodium, we recall that chlorine has a -1 charge, and sodium has a plus one charge. So that would give us sodium chloride. And as you can see, we want to continue the pattern of writing are positively charged ions first as our products in the formulas here. So now that we have our two products written out, we need to make sure we have the proper labels. So we would label nickel to sulfide with the label s because we're told in the question that this is a solid black precipitate. And this means that our second product is going to be an Aquarius product. We also want to label our reactant with the acquis label as well since they were in solution. Our last step is to make sure that our equation is balanced. And in order to balance this equation out, we want to place the coefficient of two in front of our sodium chloride so that everything is now balanced. Now that we have our fully balanced chemical equation, we need to calculate the theoretical amount of nickel to sulfide our solid precipitate that forms. So we're going to use the mass is given for each reactant to see which one produces the lowest mass of art precipitate. And that will that atom will therefore be the limiting reagent, which will determine the correct amount of nickel to sulfide are solid precipitate that forms. So what we're going to find is the mass of our nickel to sulfide formed from our first free agent, which is our Nickel to Chloride. So, NICL two. So we want to make note of the ratio between our nickel to chloride and our nickel to sulfide. And we would get this ratio from our coefficients and our balanced reaction. So looking at our nickel to chloride, we have a coefficient of one and looking at our nickel to sulfide, we have a coefficient of one as well. So we have a 1-1 ratio here and we're going to use this for our following story geometry in a second. So to begin our story geometry, we want to start off with the mass given for nickel to sulfide, which is given to us, or sorry, nickel to chloride, which is our reactant and were given a mass of 1.45 g of an I C. L. To nickel to chloride. And now we want to go ahead and using the molar mass to go from grams to moles of nickel to chloride. We're going to recall that for every atom in nickel to chloride. We get a molar mass from our periodic tables equal to 129.60 g for one mole of nickel to chloride. Now we want to go from moles of nickel to chloride into moles of our product that we're trying to find which is a precipitate nickel to sulfide. And so at this point we can plug in our molar ratio which came from our balanced equation where we said, we have for one mole of nickel to chloride. We have one mole of nickel to sulfide. And so this allows us to now counselor units of moles of nickel to chloride as well as grams of nickel to chloride, leaving us with nickel to sulfide as our product in moles. And so this is going to give us a final result equal to a value. And actually we should continue on to get this value of our product in grams. So we're going to actually extend this geometry here by utilizing molar mass of our nickel to sulfide so that we can have one mole of nickel to sulfide Equal to its molar mass for every atom and nickel to sulfide from our periodic tables of 90.76 g of nickel to sulfide. And so finally, this also allows us to get rid of moles. Which leaves us with the mass of nickel to sulfide produced, which is actually the complete way to answer this first part of our question. So this gives us a yield a theoretical yield of 1.02 g of our nickel to sulfide. And again, this is our theoretical mass produced from our first re agent, nickel to chloride. Now we want to go ahead and follow the same steps for our second re agent. So in this case we're going to find the theoretical mass of nickel to sulfide formed from our sodium sulfide. So we want to look at the ratio of sodium sulfide to nickel to sulfide. And according to our balanced reaction for nickel to sulfide or sorry, sodium sulfide, we have a coefficient of one and for nickel to sulfide, we still have a coefficient of one as well. So we have a 1 to 1 molar ratio and we're going to use that also as a conversion factor. So beginning with the mass they give us for sodium sulfide. We have a mass of 10.950 g of sodium sulfide. And we're gonna use the molar mass of sodium sulfide to get two moles of sodium sulfide. So using the periodic table, we would find that we have a molar mass of 78.04 g of sodium sulfide for one mole of sodium sulfide. Now we're going to use that molar ratio to go from moles of sodium sulfide. And sorry, they should say moles to go to molds of our product nickel to sulfide. And then lastly, we want to go from moles of nickel to sulfide. And just so this is legible. So this says moles of nickel to sulfide and we want to get two g of nickel to sulfide. So, first plugging in that molar ratio, we said from our bounced equation, we have one mole of sodium sulfide for one mole of sodium of nickel to sulfide. And then we recall from our periodic tables that for one mole of sodium or nickel to sulfide, we still have that molar mass of 90.76 g. And so now we can go ahead and cancel out our units of moles of nickel to sulfide As well as moles of sodium sulfide and grams of sodium sulfide leaving us with our mass of nickel to sulfide as our product here. And this is going to give us a value equal to 1.10 g of our nickel to sulfide. And so we can deduce that one point oh two g of nickel to sulfide is less than 1.10 g of nickel to sulfide. And so therefore because this lower mass here, 1.2 g of nickel to sulfide was produced from our sodium chloride, we would say, or sorry, our nickel to chloride. We would say that nickel to chloride is the limiting reagent And 1.02 g of nickel to sulfide is produced. And so this is going to be our final answer to complete this example. So I hope that everything I explained was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video.