An ore of barium contains BaCO3. A 1.315 g sample of the ore was treated with HC

Question

An ore of barium contains BaCO3. A 1.315 g sample of the ore was treated with HCl to dissolve the BaCO3. The resulting solution was filtered to remove insoluble material and then treated with H2SO4 to precipitate BaSO4. The precipitate was filtered, dried and found to weigh 1.074 g. What is the percentage by mass of barium in the ore? (Assume all of the barium is precipitated as BaSO4.)

Explanation / Answer

An ore of barium contains BaCO3. A 1.580 g sample of the ore was treated with HCl to dissolve the BaCO3. The resulting solution was filtered to remove insoluble material and then treated with H2SO4 to precipitate BaSO4. The precipitate was filtered, dried, and found to weigh 1.300 g. What is the percentage by mass of barium in the original sample? The initial reaction: BaCO3(aq)+2HCl(aq)---->BaCl2(aq)+H2O(l… Relative Formula Mass(RFM) of BaCO3= 208 208g= 1 mole of BaCO3 1.580g= 1.580/208 = 7.60x10^-3 moles of BaCO3 RMM of BaCl2= 219 No of moles of BaCl2 formed= no of moles of BaCO3 according to the equation = 7.60x10^-3 moles 1 mole of BaCl2= 219g 7.60x10^-3 moles of BaCl2= 219 x7.60x10^-3 = 1.6644g The 2nd reaction: BaCl2(aq)+H2SO4(aq) --->BaSO4(s)+2HCl(aq) Expected no of moles of BaSO4= no of moles of BaCl2 = 7.60x10^-3 RMM of BaSO4= 234 1 mole of BaSO4= 234g 7.60x10^-3 moles of BaSO4= 234x7.60x10^-3 = 1.7784g % barium in the original sample= (mass of BaSO4/expected mass of BaSO4) x 100% = (1.300/1.7784) x 100% = 73.1% I hope this helps! . An ore of barium contains BaCO3. A 1.640 g sample of the ore was treated with HCl to dissolve the BaCO3. The resulting solution was filtered to remove insoluble material and then treated with H2SO4 to precipitate BaSO4. The precipitate was filtered, dried, and found to weigh 1.192 g. What is the percentage by mass of barium in the original sample? molecular mass of BaSO4 = 137.33+ 32+ 16x 4= 233.33g/mol moles of BaSO4 = 1.192g/233.33 = 0.005109mol BaCl2+ H2SO4----> BaSO4 + 2HCl thus moles of BaCl2 = 0.005109mol (mole ratio of BaCl2 to BaSO4= 1:1) BaCO3 + 2HCl ---> BaCl2 + H2CO3 moles of BaCO3 reacted = 0.005109mol (mole ratio of BaCl2 to BaCO3= 1:1) molecular mass of BaCO3 = 137.33 + 12 +16x 3 = 197.33g/mol mass of BaCO3 = 0.005109mol x 197.33g/mol =1.00816g mass of Ba in BaCO3 = 137.33/197.33 x 1.00816 =0.7016g thus % mass of Ba in sample = 0.7106g/ 1.64g x 100% = 42.78% An ore of barium contains BaCO3. A 1.495 g sample of the ore was treated with HCl to dissolve the BaCO3. The resulting solution was filtered to remove insoluble material and then treated with H2SO4 to precipitate BaSO4. The precipitate was filtered, dried, and found to weigh 1.152 g. What is the percentage by mass of barium in the origin Two answers from 2 top contributors with different results - what is going on. You finally have 1.152g BaSO4 Molar mass BaSO4 = 233.3909 g/mol Molar mass Ba = 137.33 Mass Ba in 1.152g = 137.33/233.3909*1.152 = 0.6779g % Ba in original sample = 0.6779/1.495 *100 = 45.34% Now you have 2 answers that agree, So you know which is most likely correct.

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