Question

Please print.) 7. (10 pts) If 0.1800 g of impure soda ash (Na2CO) is titrated with 15.66 mL of 0.1082 M HCl (see reaction below), Na CO3(aq) + 2 HCl(aq) → 2 NaCl(aq) + H2O(l) + CO2(g) what is the percent purity of the soda ash?

Answer 1

kindly post different questions separately. please rate

of impure sodo ash- 0.1800 g Volume of Hel= 15.66ml = 0.01566 L (1000 ML -IL) concentration of Hol = 0.1082 M moles of Hel: Hol= concentration (Mly volumell) concentro 0.1082 M X 0.01566 0.01566L 1.694 X10-3 mol - N92C03 (qq+ 2HCl toy) 2 moles of Hel react with -> 2naClay! I mole of NO2003 1.694 X 16 mol Imol Imol x x 1.694 X 10-3 mole of Hel moles of No. con = moles of Na₂CO3 = react with - 8.47 810-4 mol → : moles of Nallos present 8.47 X10-4 mol 105.99 g/mol molar mawe N92C03 - mas of bure NO2CO3 = moles x molar man = 8.47 X10-4 mol X 105.99 g/mol. = 0.0898 g Percent Party of soda ash = mass of pure Naa (Oz. mans of impure No. (03 0.0898 9 x 100 0.1800g = 49.9.7. Answer