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Amixture consisting of 15.0 g ofiron(II) sulfate and 15.0 gof sodium phosphate is reacted in the following reaction: 3FeSOaaq) 2Na3- POA(aq) Fes (PO4)2s) + 3Naz= SO...

Question

Amixture consisting of 15.0 g ofiron(II) sulfate and 15.0 gof sodium phosphate is reacted in the following reaction: 3FeSOaaq) 2Na3- POA(aq) Fes (PO4)2s) + 3Naz= SOa(aq) Formula weight of FeSO4 151.91g/mol Formula weight of NazPO4 163.94 g/mol Formula weight of NazS04 119.05 g/mol Answer the following two questions:What is the maximum number of grams of sodium sulfate that can be obtained?Calculate the grams remaining of the excess reactant:

Amixture consisting of 15.0 g ofiron(II) sulfate and 15.0 gof sodium phosphate is reacted in the following reaction: 3FeSOaaq) 2Na3- POA(aq) Fes (PO4)2s) + 3Naz= SOa(aq) Formula weight of FeSO4 151.91g/mol Formula weight of NazPO4 163.94 g/mol Formula weight of NazS04 119.05 g/mol Answer the following two questions: What is the maximum number of grams of sodium sulfate that can be obtained? Calculate the grams remaining of the excess reactant:



Answers

What mass of solution containing $6.50 \%$ sodium sulfate, $\mathrm{Na}_{2} \mathrm{SO}_{4},$ by mass contains $1.75 \mathrm{~g} \mathrm{Na}_{2} \mathrm{SO}_{4} ?$

Here we have to find the mass of the solution containing 6.50% shooting himself. It by mass Contains 1.50 g and it to Sfo let the mass of the solution. Yes. Ex ground Now 6.5 of X. Graham is 1.50 grandma should himself. There are four, 6.5 Of eggs is 1.50. There are four X is 1.50 times 100 divided by 6.5. There are four of value for X. S 23.77 grounds. There are four. The mass of the solution is 23.08. Okay?

So in this reaction we have a deed iron reacting with soft fight to give so fit and how you died. All right. So, we're supposed to write balance net any equation for the reaction than a mass of sodium. Saw Fight needed to react with this mass of sodium iodide. Now again for right. Full videos on how to write balanced equations for redox reactions. Uh I would refer you to videos on such balancing of equations. Uh So, for this particular reaction. Uh This is the balanced equation for the reaction. Now for question be uh You know the massive studio cell fight. We need to start with the sodium iodide. So we will convert that to malls. So 5.0 g of said young. I did. Okay. Okay, sodium I did. And the mill amounts of sodium I did is Mhm. One 97.8. My grams per more. Mhm. Then we will compare that with the sulphites the more asia. All right. We have three moles absurd. Um Okay. So fight. Mhm. Search him suck. Right. And uh we have one more this so one more. Yeah. Okay. Of the sodium. I did. And this gives us the most of the sulphites. Which is zero point 07 58 smalls of surgeon. So fights. So the mass we will convert the most mess. So 0.7 58 most of Okay. Yeah. We use the miller mess up sodium salt. Fight, which is 1 26 light for three g In one more john's bay a move which is nine point by five g of stadium sell flights. All right. So, we started with what is known five g of that. So we convert it to move is more ratio to know the most of this. Then converted the most grams. Okay.

This is a limiting reactant problem where we have an amount for the reactant, sodium hydroxide and an amount for the reactant sulfuric acid and were asked to determine how many grams of sodium sulfate can be produced from this balanced chemical reaction. So we'll take the 60 g of sodium hydroxide, divided by the molar mass of sodium hydroxide to get mold sodium hydroxide. Then convert the mold sodium hydroxide into molds sodium sulfate with the 2 to 1 molar relationship coming from the coefficient of the balanced chemical reaction. Then when we have mold sodium sulfate, we simply multiply by the molar mass sodium sulfate to get grams sodium sulfate And 60 g sodium hydroxide will produce 165 g sodium sulfate. We'll do the same calculation with sulfuric acid, starting with the 2020g divided by its smaller mass to get moles. Then we see the 1-1 more relationship, and we'll convert moles sulfuric acid into molds sodium sulfate and then multiply by the same Mueller mass of sodium sulfate in order to get grams sodium sulfate. So 20 g sulfuric acid will produce 29.0 g sodium sulfate. This is the smaller of the two. Sophie. Uric acid is the limiting reactant and the theoretical yield, or the actual amount of sodium sulfate produced will be the smaller of these two, at 290 g

Let's calculate the number of kilograms of sodium chloride required to produce 5 kg of sodium sulfate, starting from 5 kg of sodium sulfates will convert to grams. 1 kg went to 1000 g. Moeller masses sodium sulfate is 142.1 g sodium sulfate in one more reaction. Stoke Yama Tree two moles of sodium sulfates to formals sodium chloride carry on down here one more sodium chloride, 58.5 g sodium card. Is it smaller mass and then back to kilograms, 1000 g of 1 kg and working this out? This works out to yes, 4.12 kilograms of n his sodium chloride double check this here 4.12 kg of sodium chloride that would be required.


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