Question

Almost all Nak. Salts of nitrat 12. The combustion of 0.0272 mole of a hydrocarbon produces 1.9609 H2O and 3.5927 g CO2. What is the molar mass of the hydrocarbon? 13. What is the pH of 2.9 10- M HCl(aq)? 14. What minimum mass of cobalt(II) nitrate must be added to 60.0 mL of a 0.0890 M phosphate solution in order to completely precipitate all of the phosphate as solid cobalt(II) phosphate? 2PO4 (aq) + 3 CO(NO3)2(aq) → CO3(PO4)2(5) + 6NO3 (ag)

Answer 1

11) C7H6O3 + C4H6O3 -> C9H8O4 + C2H4O2

given 2.04g of Aspirin(C9H8O4 M=180.2g/mol) is produced by 3.03g C7H6O3 (M=138.1g/mol) and 4.01g C4H6O3 (M=102.1g/mol)

we know from the reaction 138.1g of C7H6O3 reacts with 102.1g of C4H6O3

thus 3.03g of C7H6O3 will react with $102.1*3.03/138.1=2.240g$ of C4H6O3

Here C7H6O3 is the Limiting reagent

also, 138.1g of C7H6O3 produces 180.g C9H8O4

3.03g of C7H6O3 will produce $180*3.03/138.1= 3.95g$

theoretical yield = 3.95g

actual yield = 2.04g

$percent yield = actual yield*100/theoretical yield =2.04*100/3.95$ = 51.64%

12) given 0.0272 mol of hydrocarbon produces 1.9609g H2O and 3.5927g CO2

moles of H2O= 1.9609/18= 0.109mol

moles of CO2 = 3.5927/44 = 0.081mol

thus 0.0272 mol of hydrocarbon produces 0.109mol H2O and  0.081mol CO2

thus 1mol of hydrocarbon produces 4molH2O and  3molCO2

thus hydrocarbon contains 3carbon atoms and 8 hydrogen atoms. Hydrocarbon is C3H8 (propane)

molar mass=44g/mol

13) conc of HCl= $2.9*10{^{-3}}$

pH= -log[H+] = -log($2.9*10{^{-3}}$) = 2.537

14) given 60mL of 0.0890M phosphate soln.

no. of moles of phosphate soln. = 0.00534moles

from eqn, 2moles of phosphate soln are completely precipitated by 3moles of cobalt(II) nitrate.

0.00534 moles of phosphate soln will be completely precipitated by 3*0.00534/2moles of cobalt(II) nitrate.

that is 0.00801moles

molar mass of cobalt(II) nitrate = 183g/mol

mass required=183*0.00801g= 1.466g