f.5 molar volume calculations

no. of mole of CO2 = (114/1000)/24
= 0.004875 mol
NaHCO3 + HCl → NaCl + H2O + CO2
mole ratio of CO2 : NaHCO3 = 1 : 1
so, no. of mole of NaHCO3 = 0.004875 mol
mass of NaHCO3 = 0.004875 x 84
= 0.4095 g
percentage by mass = (0.4095-0.420)x100%
= 97.5%

concentration of hydrochloric acid is not used. it has no use, just to distract you. it is only useful if volume of acid is asked.

2006-10-31 23:12:43 補充：
~~sorry~~no. of mole of CO2 = 0.00475 mol, as well as no. of mole of NaHCO3.mass of NaHCO3 = 0.399 gpercentage by mass = (0.399/0.420)x100%= 95%~~SORRY~~

What is the ans. of this question ? PLEASE CHECK

To me. i think the ans should be 90%

as the equation can be resprensented in this way:

NaHCO3+HCl -- H2O+CO2+NaCl (balanced)

by the information of 114 cm3 of CO2

we can know that there are 0.00475mole of CO2 produced

coz (114/1000)/24 =0.0045 mole

thus the no. of mole of NaHCO3 = 0.0045mole (ratio by the equation)

Hence , find the mass of pure NaHCO3 = 0.0045X(23+1+12+16X3)
= 0.0045 X 84 =0.378g
thus the pencetage is

0.378/0.42 X 100% = 90%

I suspect whether i am true or not...because i haven't use the information - 2M HCl