Question

Suppose you add a solution of calcium hydroxide to a solution of sodium carbonate and a precipitate forms. You measure the total amount of the precipitate (in g) as a function of time. At time 0, 20.0 s, 40.0 s and 60.0 s you get a cumulative weight of 0.00 g, 4.82 g, 6.33 g and 8.49 g of the precipitate, respectively. What is the average rate of reaction between 40 and 60 seconds in mol/s? Enter a numerical value below without any units or symbols. If you use scientific notation, then be sure to format it using the "E" notation like "1.0E-5" and do not put spaces between the "E" and the numbers.

Answer 1

rate of reaction = change in concentration / time taken

concentration = moles = given mass / molar mass

Ca(OH)2 + Na2CO3 -------> CaCO3 (precipitate) + 2 NaOH

Molar mass of CaCO3 = 100 g/mol

[CaCO3] (in mol)   0.00        4.82 x 10^-2             6.33 x 10^-2            8.49 x 10^-2

time(s)                       0                20                             40                           60

r = 4.82 x 10^-2 / 20 = 2.41 x 10^-3 mol/s