Answer:
The correct answer is : 0.025.
Explanation:
The precipitation reaction is as follows in this procedure:
NiSO4.6H20 (aq) + Na2CO3.10H2O (aq)⇒ NiCO3 (s) + Na2SO4 (aq) + 16H2O
So, the number of moles of reactants can be found by
number of moles : mass used/ molar mass
n (NiSO4×6H2O) = (6.57 g)/(262.84 g/mol) = 0.025 mol (for Ni)
n (Na2CO3×10H2O) = (7.15 g)/(286.14 g/mol) = 0.0250 mol (for CO3).
Thus, The maximum number of moles of NiCO3 that can form is 0.025 mol.
