0.9447 g of impure NazSO4 (MW 142.04) is titrated with 42.24 mL of 0.1212 M Ba(NO3)2 according to the equation:SO4 2(aq) Bat2(aq)BaSO4s)What is the % wlw NazSO4 in ...

Challenge If the mole fraction of sulfuric acid $\left(\mathrm{H}_{2} \mathrm{SO}_{4}\right)$ in an aqueous solution is 0.325 , how much water, in grams, is in 100 $\mathrm{mL}$ of the solution?

In this question, we need to calculate the mass of magnesium sulfate that will form when we start off with one g of magnesium hydroxide and 0.605g of sulphuric asset. Now first of all we need to check if the reaction equation is balanced or not and we can see that it's not balanced. We still need to put it to in front of the water so that it's balanced on both sides, right? And now we can go ahead and do some calculations. So because the masters of both reactions are given, there is a possibility that one of them can limit the mass of magnesium self that can form in the end. So we need to identify the limiting reactant. If there is one. Before we can say what mass of magnesium sulfate will be formed in the end. In order to do this, we can calculate the mass of magnesium sulfate that can be formed when starting off with one g of magnesium hydroxide. And then just compare this mass of magnesium sulfate With the massive mechanisms offer that will form if we start off with 0.605 g of so for a gas it. So let's first calculate the mass of magnesium sulfate from that will form from um one g of magnesium hydroxide. So first of all we need to convert the mass of magnesium hydroxide to the number of moles of magnesium hydroxide. So this is going to be one g divided by the molar mass of magnesium hydroxide which is 85 point 3 2 g per more. And then we need to use the Mowlam. Uh the mole ratio um in order to determine the number of moles of magnesium hydroxide from the number of moles off. Sorry the number of moles of magnesium sulfate from the number of moles of magnesium hydroxide. So the mole ratio here is going to be 1/1. According to the strict geometric co efficiency. It's one of the one. So the number of moles of magnesium sulfate will be equal to one times the number of moles of magnesium hydroxide. Right. And then we can calculate the mess of magnesium sulfite from the number of moles of magnesium sulfate, which is equal to the number of moles of magnesium hydroxide. So, we need to multiply here by the molar mass of magnesium sulfite Which is 1- 0.36 g per mole. All right, So we need to substitute here. That's equal to one times. We started off one g of magnesium hydroxide. We need to divide that by 58.32 cramps. But more. And now we said we multiply with 120.36 g per more. And this gives a mass of magnesium sulfite That forms in this case of 2.06 g. Right now we do exactly the same. Who would do exactly the same procedure in the case of the sulphuric asset. So let's convert the massive sulphuric acid To the number of moles first. So it's 0.60 605 g divided by the molar mass of so forget asset, which is nine 8.08 g per moon. Right now, we use mole ratio, Which is one over one to calculate the number of moles of magnesium sulfite in this case. So this is 1/1. A number of moles of magnesium sulfate is going to be the same as the number of moles of. Um So for acid and from this point we can calculate the mass of magnesium sulfate by multiplying with the molar mass of magnesium sulfur, which we said is 1 20 .36 g per month. So let's substitute the values to calculate the number of models freak acid, 0.605g, divided by the molar mass of suffering acid, which is 98 08 Crimes Per Moon. And then multiply this with the molar mass of magnesium sulfate, 1 20.36 graham, sperm. All this works out two A mass of zero seven 4 to 0 .742g of magnesium sulfate. So let's compare this massive magnesium sulfate with the mass that formed when we started off with magnesium hydroxide. So clearly the second attempt gave the least amount of product. So this limits the amount of product that can be formed. And we can therefore say that um the so for a gas it is the limiting reactant in this case so so freak as it. And the mass, the maximum um mass of magnesium sulfur that can form here. Sorry this is magnesium. Yeah so fight Is equal to 0.74 two g. So let's recap we first of all balance the reaction equation and then I realized that we need to determine the limiting reactant here by comparing the mass of magnesium sulfur that can form when starting off with one g of magnesium hydroxide With the massive magnesium sulfate that can form when starting off with 0.605 g of um sulfuric acid. So we calculated the massive magnesium sulfate that could be formed in each case. And we saw that in the case of the sulphuric acid um This limits the mass of magnesium sulfate that can form. So this is the limiting reacting and the final mass of magnesium sulfate 0.742 g.

Let's first start with the definition of polarity. Since this is what we will be calculating similarity is defined as the moles of solute per liter of solution. So if we want to calculate polarity, we need to calculate the moles of solute. How can we get the moles of solute from .440 g of the salute. Well we'll use the molar mass so we'll take the grams of the salute, potassium cyanide Divide by its smaller mass, which is 97.181 g. And this will give us the moles of solute. We then divide by The volume in leaders. We were not provided the volume and leaders but we do have it in Miller leaders which we can easily convert to leaders by dividing by 1000. This will then give us moles of solute per liter solution, which is more clarity at .0133 moller.

Consider this reaction. We have mg too positive plus to which negative the maximum abound off Hydro excite. I'm exist as mg which toe? This can be calculated using ke sp expression. So substituting the values into that we have care spilling to be provided at what 0.5 10 days to put negative 11. For magnesium, the concentration is 0.1 for which negative. Who's queer? Resolving this we have which negative whole square equals 1.5 10 days to put negative 11 divided by 7.1, which gives which negative I'd concentration is 1.2 10 days to put negative five. This is the maximum amount off which negative the reaction off a niche three with water can be represented. US image three. Plus It's too gives image. Fool wants to do less, which make it do the negative. I'm produced from the next 30 must surpass to point C 2.0 10 days to plan negative five Moeller by buffering the solution through an itch for seal, then the required amount off. An urge for positive should be calculated using the expression off baby equals image food Positive which negative line and and h three the arranging this equation tau determine the concentration for and each full board do will be equal one point toe. Then these two negative for him. 0.9 Divided by Penders Toe Negative five, Which gives the answer as 1.33 mortar The mask off manage four positive required will be equal to so the mass oh, an itch for positive will be equal to 1.33 53.5 0.10 Equal tau 7.11 gram to prevent the precipitation off compound MG, which two.

So here we're looking to calculate the moles of sodium hydroxide that was feel treated with an excess of hcl, hydrochloric acid 12 N. A. H. Acquiesce H C. L. It's in equilibrium with N. A. A. C. L. Or salt. Aquarius add water, H. 20 So the moldovan uhh that's equal to 9.102 molar, multiplied by 19.85 male divide by 1000 mil. So we got 2.2 times 10 to the minus three moles. We can calculate the moles of HcL that reacts with M G. O. H. Two. So the moles of hcl, it's equal to not 0.205 mol up, multiplied by 100 mil, divide by 1000 mail. We get 1.2050 more or less the weekend. Look at the exact number of moles of HcL that reacts with M. G. O H. Two. So we have hcl is equal to not point not to 050 moles. Takeaway not point not not to 02 We got no point no 1845 moles. Okay, so we can calculate the massive M G O H two that react with our H. C. L. So we have two miles of HcL that reacts with one mole of M G O H two. If we look at the violence reaction, we have M G O H two uh in the actress condition are two hcl acquia skip's MG cl 22 H 20 So therefore we see not point not 184 or five miles of Hcl reacts with 9.225 times 10 to the minus three moles of M G O H two. Because we have a want to strike geometric relationship. We take the malls that we determined for HcL and then just divided by two. So the more massive M G O H two is 58.32 g per mole. You can calculate that from the periodic table so you can calculate the mass of mg H two. That's equal to 9.225 times 10 to the minus three moles. So the moles, we calculate it multiplied by the mola maths. What we get is not 0.538 g and so it's cabinet must present of NGO H two. In the sample, we have not 20.538 g, divide that by 9.5895 grand's multiplied by 100%. We got 91.3%.