Did I do this right? Can someone correct me Data Part I Mixture number: Mass of

Question

Did I do this right? Can someone correct me

Data Part I Mixture number: Mass of the clean dry test tube: 24.937 Mass of test tube and hydrated miature (use the whole sample):21072 Mass of test tube and anhydrous residue after 1" heating Mass of test tube and anhydrous residue after 2d heating Mass after 1" heating- mass after 2 heating Itmust be & r Mass of test tube and anhydrous residue after 3 heating Mass of water lost: (use the last mass :08152 2.2 Percent mass loss: in, the int man, 27.07t Calculations and Revults Part Mass of minture before heating: Mass of water removed: Moles of H 0 in sample: (Show colculation for credit) 21122 152 64125 -mol 10157 s 001715 m) ,0 00 11101mol Moles of BaCl 2H,O in your sample: (Show caiculation for credit) 003771 4419 Grams of BaCl 2H0 in your sample: (Show cakulatian for credit)Chet (244 26 a/mol e ol .ql g % Percent by mass BaCl 2Hg0 in your original sample: (Show cokulation) Instructor Idste & initiall:Ise lER

Explanation / Answer

Moles of BaCl2.H2O calculated by you is not correct.2 moles of water are there in 1 mol of BaCl2.H2O.This means

the number of moles of BaCl2.2H2O = 1/2 moles of H2O

=1/2 (.009725)=.0048625 moles.

So,moles of BaCl2.H2O in sample =.0048625

Now, the mass of BaCl2.H2O in sample=moles × molar mass

= .0048625×244.26=1.18772gm

So, percent by mass=

(Sample mass - mass of BaCl2.2H2O )/sample mass×100

=(2.1422-1.18772)/2.1422×100 =44.56%

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