How many grams of nico3 will be formed when a 0.300 m nicl2 solution reacts completely with 14.3 ml of a 0.191 m na2co3 solution?

0.324 grams. The balanced equation for the reaction is: NiCl2(aq) + Na2CO3(aq) = NiCO3(s) + NaCl(aq) Since the problem assumes an excess of NiCl2, we need to determine how many moles of Na2CO3 we have which will be 0.0143 * 0.191 = 0.0027313 mols And looking at the balanced equation, we can see that for each mole of Na2CO3 used, one mole of NiCO3 is produced, so we will have 0.0027313 moles of NiCO3. Now we need to determine the molar mass. So look up the atomic weights of all involved elements. Atomic weight nickel = 58.6934 Atomic weight carbon = 12.0107 Atomic weight oxygen = 15.999 Molar mass NiCO3 = 58.6934 + 12.0107 + 3 * 15.999 = 118.7011 Now just multiply the number of moles by the molar mass. 118.7011 g/mol * 0.0027313 mol = 0.324208314 g Rounding to 3 significant figures gives 0.324 grams.

The answer this would be radiative zone