Question

Data:

Mass of crucible and cover (g): 30.1500g

Mass of crucible, cover, and mixture (g): 30.6208

Mass of crucible, cover, and residue after reaction with HCl (g):
1st weighing: 30.1371g
2nd weighing: 30.1372g

Mass of unknown sample: .4708g

Mass of NaCl formed: .01285g

A. Solve the two equations and calculate the mass of NaHCO₃ and the mass of Na₂CO_3-

NaHCO₃+HCl--->NaCl+CO₂+H₂O

NaCO₃+2HCl--->2NaCl+CO₂+H₂O

B. Calculate the percent NaHCO₃ in the mixture.

C. Calculate the average percent by mass of other elements in unknown sample.

Answer 1

Mass of unknown sample: 0.4708 g

Mass of NaCl formed: 0.01285 g

and

mole of NaCl = 0.01285 g / 58.5 g / mole = 2.1966 * 10^-4 mole.

NaHCO3+HCl--->NaCl+CO2+H2O

one mole NaHCO3 produce one mole NaCl.

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

one mole Na2CO3 produce 2 mole NaCl.

thus

x + 2x = 2.1966 * 10^-4 mole

or

x = 7.322 * 10^-5 mole.

thus

mole of NaHCO3 = 7.322 * 10^-5 mole

and

mass of NaHCO3 = mole * molar mass = 7.322 * 10^-5 mole * 84.007 g/mole = 6.151 * 10^-3 g

percent NaHCO3 in the mixture = 6.151 * 10^-3 * 100 / 0.4708 = 1.306 %

and

average percent by mass of other elements in unknown sample = 100 - 1.306 = 98.694 %