Ammonia, NH₃, reacts with hypochlorite ion, OCl

Question

Ammonia, NH₃, reacts with hypochlorite ion, OCl^–, to produce hydrazine, N₂H₄. How many grams of hydrazine are produced from 115.0 g NH₃ if the reaction has a 81.5% yield?

2 NH₃ + OCl^– → N₂H₄ + Cl^– + H₂O

Accepted Answer

In this practice question, it says, ammonia, which is represented as NH 3, reacts with hypochlorite ion, OCl minus to produce hydrazine which is n 2h4. How many grams of hydrogen are produced from 115 grams of ammonia if the reaction has an 81.5% yield. Alright. So they're giving us the percent yield because it's 81.5%. Remember, percent yield equals actual yield over theoretical yield times 100. They're giving us 81 point 5, so that's 81.5 percent here. They're asking me how many grams of hydrazine are produced. So they're saying how much is produced, and what they're trying to imply here is how much do we make in real life from this experiment. We know the percent yield, so we should actually be able to determine how much of the product we actually make. Alright. So what they're asking us to solve for is actual, so that becomes our x. But I can't solve my actual yield yet because I'm still missing my theoretical yield. How do I figure out my theoretical yield? Well, to figure out a theoretical yield, we use stoichiometry. Here is our grams of given for a reactant, and if I do stoichiometry, I can figure out how much of the product I could make theoretically. So that will represent my theoretical yield. Plug that into the formula and solve for the x variable. Alright. So we're gonna have 115 grams of ammonia, and what we're gonna do is we're going to change our grams of given into moles of given. So 1 mole of ammonia, it has 1 nitrogen and 3 hydrogen. The combined mass is 17.034 grams. Remember, we're trying to figure out the grams of hydrazine. So we know we have to go all the way to grams of hydrazine in terms of the stoichiometric chart. So we're gonna go from moles of given to our moles of unknown. Remember, at this point, to go from moles of given to moles of unknown, we use the jump and we do a mole to mole comparison. For every 2 moles of this, there is 1 mole of this. So these moles cancel out. Now, finally, I'm going to say for every 1 mole of hydrazine, we need its combined mass of the 2 nitrogens and the 4 hydrogens. When we look at the periodic table and add them all up together, its combined mass is 32.052. Here, that's gonna give me 108 point oh, actually, that's gonna give me 10.82 grams of hydrazine. So take that number and plug it over here. That is our theoretical yield. So now this becomes a simple, type of algebraic question where we just have to solve for the x variable. So I'm gonna bring it over here. So it's 81.5 equals a 100 and x are both numerators, so they're it's a 100x divided by 10.82. Multiply both sides by 10.82, so these cancel out here. So when I do that, that's gonna give me 881.83 equals 100 x, divide both sides now by 100, and when I divide both sides by 100, I'm gonna see that x equals 8.8 2 grams of hydrazine, roughly. Now remember, we know that our answer here will be in grams of hydrazine, because they asked us to find grams of hydrazine here in the question. So just remember we're dealing with the percent yield formula. Reading the question carefully tells you which components of the formula you have and which one you have to solve. The wording was a little bit elusive, a little bit, cloudy, but realize they were just asking for the actual yield. Find the theoretical, plug it in, find your percent yield, plug it in, solve for actual yield, and you'll have your answer.

Struggling with GOB Chemistry?

Ammonia, NH₃, reacts with hypochlorite ion, OCl^–, to produce hydrazine, N₂H₄. How many grams of hydrazine are produced from 115.0 g NH₃ if the reaction has a 81.5% yield?
2 NH₃ + OCl^– → N₂H₄ + Cl^– + H₂O

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Struggling with GOB Chemistry?

Ammonia, NH3, reacts with hypochlorite ion, OCl–, to produce hydrazine, N2H4. How many grams of hydrazine are produced from 115.0 g NH3 if the reaction has a 81.5% yield? 2 NH3 + OCl– → N2H­4 + Cl– + H2O

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Percent Yield practice set

Ammonia, NH₃, reacts with hypochlorite ion, OCl^–, to produce hydrazine, N₂H₄. How many grams of hydrazine are produced from 115.0 g NH₃ if the reaction has a 81.5% yield?
2 NH₃ + OCl^– → N₂H₄ + Cl^– + H₂O