1) Find the moles of each element in 1.25 mol of Na2CO3.

2) Convert 2.50 mol of Na2CO3 to mass in g.

3) Convert 5.50 g of Na2CO3 to mol.

4) Covert 5.50 g of Na2CO3 to number of particles.

A.Assuming you weighed out a 2.3684 g sample of your unknown, and dissolved and diluted it as in the procedure below, and it took 35.63 mL of a 0.1025 M HCl titrant to reach the endpoint, what are the weight percents of Na₂CO₃ and NaHCO₃ in your unknown sample? Solve the simultaneous equations similar to what we did in Ch 27.

Hints:

-Set g Na₂CO₃ = X

-Eqn A: bicarbonate + carbonate = diluted weighted mass (remember, the mass is not 2.3684, you diluted it before you titrated it. Calculate the diluted g of unknown)

-Convert the g of Na₂CO₃ to mol of HCl

-Convert the g of NaHCO₃ to mol of HCl

-mol HCl = mol HCl from Na₂CO₃ + mol HCl from NaHCO₃

-Before taking the final weight percent, undo the dilution to get back to the original wt % in the unknown

-Na₂CO₃ + 2HCl ↔ 2 NaCl + H₂O + CO₂

-NaHCO₃ + HCl ↔ NaCl + H₂O + CO₂

a 2.558 g sample of na2co3, xh2o is heated to obtain anhydrous na2co3 of mass .948 g. the value of x is obtained by:

Calculate the volume of CO2 at 12.0°C and 774 mm Hg pressureobtained by treating 31.5 g of Na2CO3 with an excess ofhydrochloric acid.

the reaction is as follows;

Na2CO3 + 2HCl ---> 2NaCl + CO2 + H2O

i find the moles of the sodium carbonate, and because the moleratio btw sodium carbonate and carbon dioxide are the same, themoles used are the same formed. Then i use the ideal gas formula,converting all of what i need to use it but i still come up withthe wrong answer.. anyone please help