Find an Online Tutor Now

Yolanda J.

asked • 05/19/20

Can someone please tell me if there is an easier way to answer this question? and if I did it correctly please.

A 2.32g sample of Na2SO4*nH2 yields 1.42g Na2SO4 upon heating. What is the value of n?


Na2SO4*nH20--->Na2SO4(s)+nH20(g)


Molar Mass Na2SO4 =142 g/mol

Molar Mass H2O = 18.016 g/mol


I'm not sure if I did it correctly I kind of guessed but got the right answer in the end?- steps I did:


water loss(nH2O)= (2.32g Na2So4*nH2O) - (1.42g Na2SO4) = 0.9g nH2O


(1.42g Na2SO4)*(1mol/142 g)=0.01mol Na2SO4

(0.9g H2O)*(1mol/18.016 g)H20 = 0.050 mol nH2O



(0.050 mol nH2O)*(1 mol Na2SO4 / n H2O) = 0.01mol Na2SO4

(nH2O)(0.050 mol nH2O)*(1 mol Na2SO4) = (0.01mol Na2SO4)*(nH2O)

(nH2O)*(0.050 mol nH2O) / (0.01)= (nH2O)

[(nH2O)*4.995 mol H20] / (nH2O) = (nH2O)/(nH2O)

=5 mol H2O



1 Expert Answer

By:

Still looking for help? Get the right answer, fast.

Ask a question for free

Get a free answer to a quick problem.
Most questions answered within 4 hours.

OR

Find an Online Tutor Now

Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.