What's 11.8 M NiCl2? The problem says M NiCl2 is 0.258. You must mean 11.8 mL.
mols Na2CO3 = M x L = 0.153 x 0.020 = ?
mols NiCO3 formed = same since there is 1 mol NiCO3 for 1 mol Na2CO3.
grams NiCO3 = mols NiCO3 x molar mass NiCO3.
How many milliliters of 0.258 M NiCl2 solution are needed to react completely with 20.0 mL of 0.153 M Na2CO3 solution? How many grams of NiCO3 will be formed? The reaction is
Na2CO3(aq) +NiCl2 (aq)----> NiCO3(s)+2NaCl
I solved for the 1st part, 11.8 M NiCl2
I am not sure where to begin to solve for grams.
2 answers
But note that I get 11.86 mL which I would round to 11.9 mL to three significant figures.