Question

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A 1.470 g sample of an organic acid containing C, H and was combusted to determine the empirical formula. The combustion produced 3.810g of carbon dioxide (44,01 g/mol) and 1.820g of water(18.02 g/mol). Determine the number of moles of carbon in the compound, C- 12.01 g/mol H-1.01 g/mol 0-16.00 g/mol

Answer 1

get man of Cand H We com 7 From the amount of CO ₂ we can -C in the compound; from H₂o we can can get 1 in the compound. By subs tracting in the compound present man of find oxygen present molan man of CO2 = 44.01g/mot. MM of C = 12.01g/mol in 49001g Co, there is 12.01g 3.8109 (MM) C. So, in there is 12:0) 44.01 *3.819 =1.03979 C mole of C- 1.03974 12.01 glmol = 0.0866 mol C

So, number of moles of Carbon in compound 0.0866 mol AD Balancing the equation means making number of each element same in bothsides. Potanium phosphate = kg poy Lead (11) nitrate = Pb(NO3)2 the product is lead phosphate = Pb (PO4)2 and potanium nitrate= KNO3 : Chemical equation K₃PO4 + Pb(NO3), → P13(PO4)2 + (NO3 1 balancing the equation 2K3 P04 + 3Pb(NO3), = Pb3 (904), +6 KNO3 formula of lead phosphate = Pb₃(PO4)2 balanced equation Coefpicient of potanium nitrate It is a double displacement reaction when cations and anions switch between two reactants to form product. = 6 new AB + CD AD + BC 3- + hore A = Kt , B=P04 27 Pb D = NO3 ; C =