Suppose you add a solution of calcium hydroxide to a solution of sodium carbonate and a...

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Suppose you add a solution of calcium hydroxide to a solution ofsodium carbonate and a precipitate forms. You measure the totalamount 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 isthe average rate of reaction between 40 and 60 seconds in mol/s?Enter a numerical value below without any units or symbols. If youuse scientific notation, then be sure to format it using the \"E\"notation like \"1.0E-5\" and do not put spaces between the \"E\" andthe numbers.

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3.6 Ratings (459 Votes)

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)Â Â 0.00Â Â Â Â Â Â Â 4.82 x 10^-2Â Â Â Â Â Â Â Â Â Â Â Â 6.33 x 10^-2Â Â Â Â Â Â Â Â Â Â Â 8.49 x 10^-2

time(s) Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 20Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 40Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 60

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

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