posted by Lisa .
A thoroughly dried 1.281g sample of Na2SO4 is exposed to the atmosphere and found to gain 0.383g in mass.
What is the percent, by mass, of
Na2SO4*10 H2O in the resulting mixture of anhydrous Na2SO4 and the decahydrate
I can not be sure about this one, but I will give it a try.
0.383g is H20
since you know that 10 moles of H20= 1 mole of Na2SO4, then 0.383g of H20*(1 mole of H20/18g of H20)=moles of H20 in sample,
moles of H20 in sample*(1 mole of Na2SO4/10moles of H20)= moles of Na2SO4 that needs to react with 0.383g of H20.
moles of Na2SO4 *(142.0 g of Na2SO4/1 mole of Na2SO4) = g of Na2SO4 needed to react with H2O
(g of Na2SO4 needed to react with H2O + 0.383g of H20)/(0.383g + 1.281g)*100= % by mass
I would do this through another route;I agree with Devron with 0.302g Na2SO4 used and the percentage comes out the same but uses different approaches.. Take you pick.
1.281 = g Na2SO4 initially. Some of that reacts to form Na2SO4.10H2O and some of it remains as Na2SO4. We need to find the two parts; i.e., the amount remaining intact and the amount used to make Na2SO4.1OH2O.
(Note: mm stands for molar mass.)
How much will the Na2SO4.10H2O weigh? That will be 0.383 x (mm Na2SO4.10H2O/mm10H2O) = 0.383 x (322.196/10*18.015) = about 0.685g Na2SO4.10H2O.
How much Na2SO4 did that use? That's
0.685 x (mmNa2SO4/mmNa2SO4.10H2O) = 0.685 x (142.043/322.196) = 0.302g.
Therefore, the 1.281 g sample of Na2SO4 can be divided into two pieces.
a. 0.302g for conversion to Na2SO4.10H2O
b. 0.979g that stays unchanged.
Therefore, the total mass when we finsih is
0.979g from Na2SO4 not changed.
0.302g Na2SO4 changed to Na2SO4.10H2O where it weighed 0.685g.
Thus total weight is
0.685(decahydrate) + 0.979(anhydrous) = 1.664g total.
% decahydrate = ( 0.685/1.664)*100 = ?