Hydrogen and chlorine react to yield hydrogen chloride: H2 + Cl2 ...

Question

Hydrogen and chlorine react to yield hydrogen chloride: H2 + Cl2 ¡ 2 HCl. How many grams of HCl are formed from reaction of 3.56 g of H2 with 8.94 g of Cl2? Which reactant is limiting?

Accepted Answer

Hi everyone for this problem. It reads magnesium carbonate is produced from the combination of magnesium oxide and carbon dioxide. A reaction vessel that initially contains 6.25 g of magnesium oxide was charged with a 6.25 g of carbon dioxide, magnesium oxide and carbon dioxide react until one is completely used up. Identify the reactant that remains after the reaction goes to completion. Okay, So the reactant that remains is going to be our excess reactant. Okay. And that is what we're trying to figure out out of our two reactant, magnesium oxide and carbon dioxide. Which one is going to remain after the reaction goes to completion. And in order for us to do this, we're going to need to know the amount of moles of product that is produced from each of our reactant. And from that, we'll see which one is our limiting reactant and which is our excess reactant. So, let's go ahead and get started. So, we have to react ints here. Alright, so let's figure out how much moles of magnesium carbonate is going to be produced by each. So, let's start off with our magnesium oxide. Okay, so we want to figure out the moles of magnesium oxide from the moles of magnesium carbonate from magnesium oxide. So, we'll start off with our amount of magnesium oxide we start with. So we have 6.25 g of magnesium oxide. Okay, So we want to go from grams of magnesium oxide to moles of magnesium carbonate. This is our goal. So, let's go ahead and use dimensional analysis here to create that road map. All right. So, starting with our grams of magnesium oxide, we're going to want to go from grams of magnesium oxide, two moles of magnesium oxide. That way we can use our multiple ratio of magnesium oxide to magnesium carbonate. So, in one mole of magnesium oxide, our molar mass based off of our periodic table is 24.305 g of magnesium oxide. Alright, so we'll cancel our units as we go. So our grams of magnesium oxide cancel. And now we're in moles of magnesium oxide. We want to go from moles of magnesium oxide, two moles of magnesium carbonate. So, we need a multiple ratio. Alright. And we're going to refer to our reaction above to look at that multiple ratio. So, looking at our reaction for every one mole of magnesium oxide consumed, one mole of magnesium carbonate is produced. That's our multiple ratio. That will put in now. So, for every one mole of magnesium oxide consumed, There is one mole of magnesium carbonate produced. And we're looking at the coefficients here. Okay, so we can see here there's a one as a coefficient for everything on our reaction. All right. So, as you can see now are moles of magnesium oxide cancel. And we're left with moles of magnesium carbonate, which was our goal. So, let's go ahead and do this calculation. When we do this calculation, we get an answer of 0. moles of magnesium carbonate produced from 6.25 g of magnesium oxide. Alright, so that was step one. Now we need to do the same thing for our carbon dioxide. We want to know how many moles of magnesium carbonate is going to be produced from our 6.25 g of carbon dioxide consumed. So we're going to do the same thing that we just did. But using our carbon dioxide as our starting. Okay, so we want to go from g of carbon dioxide, two moles of magnesium carbonate. Okay, so in one more of carbon dioxide There is what's our molar mass? 12.01g of carbon dioxide. Alright, and we did that so that our grams of carbon dioxide cancel. So now we're in moles of carbon dioxide, we want to go from moles of carbon dioxide to moles of magnesium carbonate. And we'll use the mole to mole ratio based off of our reaction, which is a 1 to 1 ratio. So for every one mole of carbon dioxide consumed, We have one mole of magnesium carbonate produced. Alright, so our moles of carbon dioxide cancel. And we're left with moles of magnesium carbonate will do the calculation and we get 0.5- moles of magnesium carbonate is produce From 6.25 g of carbon dioxide. Alright, so what do these two values mean? So we see that one is larger than the other. So what that means is the smaller number or the smaller amount is going to be our limiting reactant? Okay, this is the one that's going to be completely used up. Okay. Are smaller amount is going to be the one that's completely used up. And so that means the larger amount is going to be the one that is in excess. Okay, So are magnesium oxide is our limiting reactant and the carbon dioxide is our excess reactant. So the question asked us to identify the reactant that remains after the reaction goes to completion. And so our answer for this problem is our excess reactant is going to be carbon dioxide. All right, So let's go ahead and write that down here. Carbon dioxide equals our excess reacting. Alright, so that's the end of this problem. I hope this was helpful.

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