Question

given: 2 NiCl3(aq) + 3 Na2S(aq) --> Ni2S3(s) + 6 NaCl(aq) If necessary use Mm (g/mol)...

given:

2 NiCl3(aq) + 3 Na2S(aq) --> Ni2S3(s) + 6 NaCl(aq)

If necessary use Mm (g/mol) = 165.04 78.05 213.59

If 70.0 mL of a 0.400 M NiCl3(aq) is mixed 40.0 mL of a 0.800 M Na2S(aq), how many moles of the excess reactant are leftover?

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