Chemistry help??

Question

1) What volume of 3 M HCl would be required to react with 5 g of a mixture that is 50% by mass sodium carbonate and 50 % sodium hydrogen carbonate?

2) The NaCl was not dried suffieciently- would the percent yield be higher or lower than the theoretical yield?

3) If small portions of NaCl were lost to spattering during the evaporation of the solvent in this experiment, would the percentage yield be higher or lower? Explain

 

Answer 1

(1)

5g of mixture consist of 50% of Sodium Chloride and 50% of Sodium Hydrogen Carbonate

50% of 5g = 2.5 g

There is 2.5 g of Sodium Carbonate and 2.5g of Sodium Hydrogen

Molecular Weight of Sodium Carbonate(Na2co3)  is

= (2 * atomic weight of Sodium) + (1 * Atomic Weight of Carbon) + (3 * Atomic Weight of Oxygen)

= (2 * 23) + 12 + (3 * 16) = 106

Chemical Equation when Sodium Carbonate reacts with HCl

Na2CO3 + 2 HCl → 2 NaCl + CO2 + H2O

Number of Moles of HCl  = (2.5/106) / (1/2) =  0.047 moles of HCl is Required

 

Molecular Weight of Sodium Carbonate(NaHco3)  is

= (1 * atomic weight of Sodium) + (1 * atomic Weight of Hydrogen) + (1 * Atomic Weight of Carbon) + (3 * Atomic Weight of Oxygen)

= (1 * 23) + 1 + 12 + (3 * 16) = 84

Chemical Equation when Sodium hydrogen Carbonate reacts with HCl

NaHCO3 + HCl → NaCl + CO2 + H2O

Number of Moles of HCl  = (2.5/84) / (1/1) =  0.029 moles of HCl is Required

 

Total Number of HCl is 0.047 + 0.029 = 0.076 moles of HCl.

We know that n = cV

V = n / c = 0.076 / 3 = 0.02533 = 25.33 ml

Volume of the HCl is 25.33 ml.

Answer 2

(3)

The Theoretical Values are always higher than the Practical values because the values calculated theoretically are ideal conditions but when it comes to a practical values there are many factors effecting.

If your NaCl is spattered due to the evaporation of the Solvent then this would effect the values.

So Percent would yield to lower value.