Na2SO4 * xH2O, the hydrated salt is overheated and the anhydrous salt thermally decompose, one product being a gas; will the reported value of "x" be too high, too low on the unaffected? explain

Md Ariful Islam

0 like 0 dislike
1 views
Share with your friends

1 Answers

Md Ariful Islam

Call
Na2SO4*xH2O = (dehydratation) = Na2SO4 + xH2O.x = n(H2O)/n(Na2SO4*xH2O), where n(H2O) and n(Na2SO4*xH2O) the number of moles of the water and of the hydrated salt respectively.The sodium sulfate exists mostly in the dehydrate form: Na2SO4*10H2O, so x would be equal to 10.
0 like 0 dislike
1 views
Talk Doctor Online in Bissoy App
<ol><li> A solution is made by dissolving 125 g Na2SO4 in 65.0 g water. Calculate the mole fraction of the solute and the solvent. (MM Na2SO4 = 142 g/mole; water = 18 g/mole)</li></ol>

2NaOH+H2​SO4​→Na2​SO4​+2H2​O= 65/18 = 3.61moles125/142 = 0.88MTotal moles = 4.91H2O = 3.61/4.91 = 0.74Na2SO4 = 0.88/4.91= 0.179

1 Answers 1 views
A 3.211g sample of gypsum, a hydrated salt of magnesium sulphate is heated. The mass of the anhydrous MgSO4 is 2.539g. Calculate the amount of water released during heating.

Amount of water released during heating=5.74g

1 Answers 1 views
A 122.74 g of FeSO4•xH2O contains 77.04 g of FeSO4, calculate the percent by mass of water in the hydrate and record the value of x in the short answer box.

H2O= 122.74-77.0445.7 g% = 45.7/122.74×100 = 37.23%Molar Mass of H2O = 18.015x= 45.7/18.015x= 2.54

1 Answers 1 views
13.4 g of a sample of unstable hydrated salt: Na2SO4. xH20 was found to contain 6.3g of water. Determine the number of water of crystallization

The molar mass of the salt is m(Na2SO4* xH2O) = 2*23+32+16*4 + x (2 + 16) = 142 + x*18 [g/mol] Then let's construct the proportion: 13.4 / 6.3 =...

1 Answers 1 views
A student wants to prepare 250 mL of a 0.035 mol L-1 solution from the primary standard, anhydrous sodium carbonate. Calculate the mass of anhydrous sodium carbonate that is needed.

n(Na2CO3) = C(Na2CO3) × V(Na2CO3) = 0.035 mol L-1 × 0.25 L = 0.00875 mol.m(Na2CO3) = n(Na2CO3) × M(Na2CO3) = 0.00875 mol × (2×23 + 12 + 3×16) = 0.9275...

1 Answers 1 views
Which is bigger, hydrated Na+ or hydrated H+?

The molar mass of NaOH and H2O is (23 + 16 + 1) = 40 g/mol. and (2*1 + 16) = 18 g/mol. Thus, if we understood your question correctly,...

1 Answers 1 views
a student heated 21.5 g pure hydrated calcium sulfate ,CaSO4 x H2O .17.0 g anhydrous calcium sulfate was formed .Determine the molecular formula for this hydrated salt

CaSO4•H2O-&gt;(t) CaSO4+H2Om(CaSO4)=17gm(CaSO4•xH2O)=21.5gM(CaSO4•xH2O)=m(CaSO4•xH2O)*M(CaSO4)/m(CaSO4)=21.5*136/17=172g/molx=(M(CaSO4•xH2O)-M(CaSO4))/18=(172-136)/18=2 CaSO4•2H2O

1 Answers 1 views
2.0g of a hydrated salt XnH2O was heated to a constant mass 1.09g of the anhydrous salt X was left.given that the relative formula mass of X=152 find n

M (H2O) = 18nm (H2O) = 2.0 - 1.09 = 0,91 gM (X) = 152Ratio:1.09/152 = 0.91/18nn = 7

1 Answers 1 views
The formula for a hydrate of magnesium sulfate is MgSO4•xH2O. If 1 mol of anhydrous magnesium sulfate is 48.82% of the mass of 1 mol of the hydrate, what is the value of x?

M(MgSO4) = 120.366 g/mol;M(H2O) = 18.015 g/mol;Proportion:120.366 g (mass of 1 mole of MgSO4) — 48.82 %Total mass of 1 mole of MgSO4•xH2O — 100 %M(MgSO4•xH2O) = 246.55 g/mol;M(MgSO4) +...

1 Answers 1 views
If the endpoint is surpassed in a back titration with NaOH, will the reported amount of antacid in the sample be too high or too low? Explain.

The reported amount of antacid will be too low because if you added excess, there would be more subtracted from the acid used. The excess NaOH subtracted from the acid...

1 Answers 1 views
X