Question

im doing a dry lab, can anyone check over problems:1-4 see if i did it correct ? thank you

Name Part A Data Table Mass (9) 46.0076a 45.68259 0.3251 45.8370g 45.68759 0.1551 1. Mass of dean, dry evaporating dish and watch 45.6825. glass 2. Mass of evaporating dish, watch glass, and 910 00 Na.Co 3. Mass of Na2CO 10.3251 4. Mass of evaporating dish, watch glass, and Nadi4545210 residue after drying 5 .Experimental mass of NaCl(sodium chloride) 10.1551 6. Theoretical mass of NaCl (show work below) 1011920 7. Percent yield (show work below) 86.55% Calculation of theoretical mass of NaCl (Show work) MOLAR MASS NA, CO - =(23x2)+ 12 + 110x 3) 1 467 12+ 48. = 100g 0.3251 Na. Col I N.CO₂ Ilmo NaCl 158.45 g Nag MOLAR MASS Naci: - TOGNA, CO3 Timol Na, Co/1 mol NaCl = 23 + 35.45 = 58.459 10.1792 g NaCl Calculation of percent yield of NaCl (Show Work) 0.1551 0. 1792 x 100 = 186.55%

Answer 1

PART A

Mass of Na2CO3 = 0.3251 g

Mass of NaCl = 0.1551 g

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

Mole of Na2CO3 = weight/ molar mass = 0.3251 / 105.988 = 0.00367 moles

Molar ratio of Na2CO3 : NaCl = 1: 2

.: Moles of NaCl = 0.00367 $\times$ 2 = 0.006134 moles

Mass of 0.006134 moles of NaCl =0.006134  $\times$ 58.44 = 0.3584 g

Thus the Theoretical mass of NaCl = 0.3584 g

Percentage yeild = (Obtained experimental mass/ theoretical mass) $\times$ 100 = (0.1551 / 0.3584) $\times$ 100 = 43.27 %

PART B

Mass of NaHCO3  = 0.3498 g

Mass of NaCl = 0.2403 g

NaHCO3 + 2HCl --> NaCl + H2O + CO2

Mole of NaHCO3 = weight/ molar mass = 0.3498 / 84.01 = 0.004163 moles

Molar ratio of NaHCO3 : NaCl = 1: 1

.: Moles of NaCl = 0.004163 moles

Mass of 0.004163 moles of NaCl = 0.004163 $\times$ 58.44 = 0.2432 g

Thus the Theoretical mass of NaCl = 0.2432 g

Percentage yeild = (Obtained experimental mass/ theoretical mass) $\times$ 100 = (0.2403 / 0.2432) $\times$ 100 = 98.80 %

Q.3.  Mass of Na2CO3 = 0.3251 g

Moles of Na2CO3 = 0.00367 moles

Na2CO3 : CO2 = 1: 1

Moles of CO2 = 0.00367 moles

Mass of 0.00367 moles of CO2 = 0.00367 $\times$ 44 = 0.16148 g

Q.4.  Mass of CaCO3 = 6 g

Moles of CaCO3 = 6 $\div$ 100.0869 = 0.0599 moles

CaCO3 : H2O= 1: 1

Moles of H2O = 0.0599 moles

Mass of 0.0599 moles of H2O =0.0599  $\times$ 18 = 1.0782 g