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Problem 85

Calcium nitrate and ammonium fluoride react to fo…

07:01
University of Maine
Problem 84

Aluminum nitrite and ammonium chloride react to form aluminum chloride, nitrogen, and water. How many grams of each substance are present after 72.5 $\mathrm{g}$ of aluminum nitrite and 58.6 $\mathrm{g}$ of ammonium chloride react completely?

Answer

12.3 g aluminum nitrite, 0 g ammonium chloride, 48.6g aluminum chloride, 30.7g nitrogen, 39.5g water


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Video Transcript

given a chemical reaction and initial amount of reactant, we can predict how much of each product will form, as well as how much of any reactant remains. The first step is to make sure that you have a complete balanced equation, So if you're given the words, you need to find the formulas for everything. So aluminum nitrate is an Ionic compound made up of the aluminum ion, which has a plus three charge in the nitrate ion, which is a poly atomic ion with a minus one charge to make a neutral compound. The formula is a L and no to three. Ammonium chloride is another Ionic compound made up of the eye on NH for the ammonium ion and C L minus. For every positive, there's a negative. So the formula for ammonium chloride is NH four CEO. They react to form aluminum chloride and I on a compound made of the aluminum ion and the chlorine I, on which, when they combined to make a neutral compound, forms ales C l three plus nitrogen, which is an element and which exists as end to in most chemical reactions plus water, which is H 20 once we have the skeletal equation. We then balance the equation by changing any coefficients. If we look at the number of chlorine atoms on this side, there's one Corinne, and on this side there are three. So we multiply the ammonium chloride by three. This gives us three. Nitrogen is here as well as three night regions here, or a total of six nitrogen, so we won't apply the end to buy three. There's one aluminum on both sides. There are six oxygen's on this side, so we multiply the water by six, giving us 12 hydrogen ins and 12 hydrogen. Now that we have the complete balanced equation, we can determine how much product will form. In this case, we have 72.5 grams of the aluminum nitrate nitrate and 58.6 grams of the ammonium chloride. Since we have two quantities, we have to find what's called the limiting re agent, and the limiting re agent is the initial reactant that produces thes smaller amount of product. I can choose any product on the side. In this case, I'll start with the aluminum chloride. So if I have this many grams of aluminum nitrate The first step is to always change two moles, and we do this by using the molar mass, which will find using the periodic table. Once we have moles of aluminum nitrate, we can change two moles of product to go from one substance to another. In the balance equation, we used the coefficients or what's called the mole ratio, and then we'll do the same thing for ammonium chloride, going from grams two malls using the molar Mass and then going to the moles of product. So if I have my aluminum nitrate, I confined the molar mass of that. Using the periodic table, I find the molar mass of the aluminum 26.982 grams, plus three times the molar mass of nitrogen, plus six times the molar mass of oxygen or 100 and 64 0.997 grams. I also know from the balanced equation that for every one mole of aluminum nitrate, one mole of aluminum chloride will be produced because the ratio is 1 to 1 based on the coefficients. So if I have 72 0.5 grams of aluminum nitrate, I first changed two moles by dividing by the molar Mass. And then I changed two moles of aluminum chloride, Or I should form zero point for 39 moles of aluminum chloride. We go through the same calculation with the amount of ammonium chloride. 58 0.6 grams. NH four c l changing by using the molar mass, which I found again using the periodic table taking the molar mass of nitrogen plus four hydrogen Z plus chlorine for Moeller math off 53 0.498 grams and looking at the balanced equation, I can see that for every one mole of ammonium chloride, I'll produce three moles. Sorry for every three moles of ammonium chloride, I'll produce one mole of aluminum chloride. So by dividing 58.6 by 53.498 by three, I see that this produces 0.365 moles of aluminum chloride. Because this is the smaller quantity. That means that ammonium chloride is the limiting re agent. And so what that means is that when the reaction is completed, there will be none left. So if we want to know how much remains how many grams, there are zero grams remaining at the end of the reaction, find out how many grams of each product forms. We use the number of moles formed by the living re agent. So if I have this many moles of aluminum chloride, I confined how many grams that's equivalent by using the molar mass. So 0.365 moles of aluminum chloride. And in this case we multiply by the molar mass, which I find using the periodic table and each one of the atomic masses. 133 went 332 grams for everyone mall, so I should produce 48 0.6 grams of aluminum chloride. Find out how much nitrogen is formed. You start off with the moles of aluminum chloride, which we can then change two moles of nitrogen using the mole ratio from the balanced equation. And then we changed two grams nitrogen using the molar mass of nitrogen. So point 365 moles of aluminum chloride and then looking at the balanced equation, I can see that there are three moles and to for every one mole of a l. C. L three and the molar mass of end to is two times the molar mass of end or 28.14 grams Permal. So this gives me 30.7 grams of nitrogen to find out how much water forms the same process, starting again with 0.365 moles of aluminum chloride using the mole ratio, there are six moles of water for every one mole of aluminum chloride and then multiplying by the molar mass of water or 39.5 grams of water. Finally, we can look at how much of the excess re agent or the re agent that was not the limiting reactant, which in this case, if the aluminum nitrate so I have still the amount of product that will form or 0.0 I 0.365 moles of aluminum chloride, and I want to find out how many grams of aluminum nitrate are equivalent. I know. So I use my more issue, which is for every one mole of aluminum nitrate. There's one mole of aluminum chloride multiplied by the molar mass of aluminum nitrate, which is 100 and 64.997 grams. For every mole, this is 60 went to grams, so that's how much should be used up and remember initially I had 72.5 grams, so we then subtract those two numbers, or 12.3 grams of aluminum nitrite.