Calculate the mass of the white solid calcium carbonate that forms when 250.0 mL

Question

Calculate the mass of the white solid calcium carbonate that forms when 250.0 mL of a 0.100 M calcium nitrate solution is mixed with 200.0 mL of a 0.150 M sodium carbonate solution. 1st, write out the balanced chemical reaction:calculate the number of moles of BOTH calcium nitrate and sodium carbonate: determine which solution is the “limiting reactant” _________________________ based on this, how many moles of product, white solid calcium carbonate, do you expect to form: _________________________ , determine the mass of calcium carbonate you expect to form from the reaction:

Explanation / Answer

balanced chemical equation :

Ca(NO3)2 + Na2CO3   ---------------> CaCO3 + 2 NaNO3

b)

moles of Ca(NO3)2 = 250 x 0.1 / 1000 = 0.025

moles of Na2CO3 = 200 x 0.15 / 1000 = 0.03

c)

Ca(NO3)2 + Na2CO3   ---------------> CaCO3 + 2 NaNO3

   1                     1

0.025             0.03

limiting reagent is Ca(NO3)2.

c)

moles of product = 0.025 mol

mass of product = 2.5 g

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