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You have remained in right site to begin getting this info. get the 11 2 Practice Problems Continued Answers belong to that we present here and 0.549 moles Ni2(CO3)3 x 297.41g = 163.27 g 1 moles Ni2(CO3)3 Solving from H2CO3 3.45g x 1 mole H2CO3 = 0.0556 moles H2CO3 62.03g 0.0556 moles H2CO3 x 1 moles Ni2(CO3)3 = 0.01854 moles Ni2(CO3)3 3 moles H2CO3 0.01854 moles Ni2(CO3)3 x 297.41g = 5.514 g 1 mole Ni2(CO3)3 Ni is the excess reagent H2CO3 is the limiting reagent 6. Use to CFU. Notes on Isotopes and Atomic Mass (p's 2 and 3) - a BBC Bitesize revision site. This is page 2, continues on page 3 Isotope simulation, PhET, Uni. of Colorado. - click "abundance in Nature" and "Symbol" buttons. Set questions for students to list the symbols and abundance of isotopes of certain elements. 11-4 Practice (continued) Form K Find the volume of each cylinder in terms of π and to the nearest tenth. 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