Question

2.5 g of sample containing Na2Co3, NaHCo3 and some non volatile impurities on gentle heating loses 12% of its weight. Residue is dissolved in 100 ml of water and it's 10 ml is portion required 15 ml of 0.1M solution of BaCl2 for complete precipitation of the carbonates. Determine mass percentage of Na2No3 in the original sample.

Answer 1

Answer:approximate 6.3%

Explanation:

Answer 2

Answer :

42.4%Na2CO3

Solution :

Weight loss is due to conversion fo NaHCO3 into Na2CO3:31g weight is lost per mole of NaHCO3

⇒0.3gwt. loss from 0.331 mol of NaHCO3 producing 0.362 moles of Na2CO3

Total moles of carbonate =15×10−3

implies Moles of carbonate in original sample =0.015−3620=0.01

Mass of Na2CO3 in original sample =1.06⇒42.4%Na2CO3